diff options
| author | nfenwick <nfenwick@pglaf.org> | 2025-03-09 07:43:48 -0700 |
|---|---|---|
| committer | nfenwick <nfenwick@pglaf.org> | 2025-03-09 07:43:48 -0700 |
| commit | 8d5144568d0b55e3f8f0675b6d0d40381502e2c9 (patch) | |
| tree | d027f6c8e284a3bd918d5dbbac77cd41714a00fe /40175-h | |
| parent | 90cd347eeacca62564189c1e60d5fc4a854adacd (diff) | |
Diffstat (limited to '40175-h')
| -rw-r--r-- | 40175-h/40175-h.htm | 980 |
1 files changed, 278 insertions, 702 deletions
diff --git a/40175-h/40175-h.htm b/40175-h/40175-h.htm index 0e84884..deaef67 100644 --- a/40175-h/40175-h.htm +++ b/40175-h/40175-h.htm @@ -2,7 +2,7 @@ "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> <html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en"> <head> - <meta http-equiv="Content-Type" content="text/html;charset=iso-8859-1" /> + <meta http-equiv="Content-Type" content="text/html;charset=UTF-8" /> <meta http-equiv="Content-Style-Type" content="text/css" /> <title> The Project Gutenberg eBook of Physics, by Willis E. Tower, M. Sci. (Univ. Of Illinois). @@ -212,49 +212,7 @@ em.gesperrt </style> </head> <body> - - -<pre> - -The Project Gutenberg EBook of Physics, by -Willis Eugene Tower and Charles Henry Smith and Charles Mark Turton and Thomas Darlington Cope - -This eBook is for the use of anyone anywhere at no cost and with -almost no restrictions whatsoever. You may copy it, give it away or -re-use it under the terms of the Project Gutenberg License included -with this eBook or online at www.gutenberg.org/license - - -Title: Physics - -Author: Willis Eugene Tower - Charles Henry Smith - Charles Mark Turton - Thomas Darlington Cope - -Release Date: July 9, 2012 [EBook #40175] - -Language: English - -Character set encoding: ISO-8859-1 - -*** START OF THIS PROJECT GUTENBERG EBOOK PHYSICS *** - - - - -Produced by Anna Hall, Albert László and the Online -Distributed Proofreading Team at http://www.pgdp.net (This -file was produced from images generously made available -by The Internet Archive) - - - - - - -</pre> - +<div>*** START OF THE PROJECT GUTENBERG EBOOK 40175 ***</div> <h1>PHYSICS</h1> @@ -1077,7 +1035,7 @@ measure.</p> <p>The standard unit of <i>mass</i> is the <i>kilogram</i>. It is the mass of 1 liter of pure water at the temperature of its -greatest density, 4°C. or 39.2°F.</p> +greatest density, 4°C. or 39.2°F.</p> <p>The three principal units of the metric system, the <i>meter</i>, the <i>liter</i>, and the <i>kilogram</i>, are related to one another @@ -1145,7 +1103,7 @@ standard units may be given thus: <i>first</i>, the liter is a cubic decimeter, and <i>second</i>, the kilogram is the mass of a liter of water. (See Fig. 5) Since the liter is a cubic decimeter, the length of one side is 10 cm. The liter therefore -holds 1000 ccm. (10 × 10 × 10). Therefore, 1 liter +holds 1000 ccm. (10 × 10 × 10). Therefore, 1 liter = 1 cu. dm. = 1000 ccm. and since 1 liter of water has a mass of 1 kg. or 1000 g., then 1000 ccm. of water has a mass of 1000 g., or <i>1 ccm. of water has a mass of 1 g.</i></p> @@ -1257,8 +1215,8 @@ so that 100,000,000 molecules (a number nearly equal to the population of the United States) pass out every second, it would take, not minutes or hours, but nearly 9000 years for all of the molecules to escape. The number of -molecules in 1 ccm. of air at 0°C. and 76 cm. pressure has -been calculated by Professor Rutherford to be 2.7 × 10<sup>19</sup>. +molecules in 1 ccm. of air at 0°C. and 76 cm. pressure has +been calculated by Professor Rutherford to be 2.7 × 10<sup>19</sup>. It is evident that such minute particles cannot be seen or handled as <i>individuals</i>. We must judge of their size and action by the results obtained from experiments.</p> @@ -1334,7 +1292,7 @@ opposite directions through the walls of <i>C</i>, the hydrogen goes in much faster than the air comes out. In consequence it accumulates, creates pressure, and drives down the water in <i>W</i> and out at <i>J</i>. On removing <i>B</i>, the hydrogen within the porous cup comes out much -faster than the air reënters. This lessens the pressure within, so +faster than the air reënters. This lessens the pressure within, so that air rushes in through <i>J</i>. This experiment demonstrates not only the fact of molecular motion in gases but also that molecules of hydrogen move much faster than those of air. (This experiment @@ -2367,10 +2325,10 @@ side?</p> equals the product of the <i>area</i> of the surface, the <i>depth</i> of the liquid and its weight per unit volume, or using the formula, <i>F = Ahd</i>. To compute the downward force against the bottom we have the -area, 9, depth, 4, and the weight 62.4 lbs. per cubic foot. 9 × -4 × 62.4 lbs. = 2246.4 lbs. To compute the force against a side, the +area, 9, depth, 4, and the weight 62.4 lbs. per cubic foot. 9 × +4 × 62.4 lbs. = 2246.4 lbs. To compute the force against a side, the area is 12, the average depth of water on the side is 2, the weight -62.4, 12 × 2 × 62.4 lbs. = 1497.6 lbs.</p></div> +62.4, 12 × 2 × 62.4 lbs. = 1497.6 lbs.</p></div> <h4>Important Topics</h4> @@ -2499,7 +2457,7 @@ depth.</span> <p>If now a cubic centimeter of water be poured upon <i>AB</i> it will raise the level 1 cm., or the head of water exerting pressure upon <i>CD</i> becomes -11 cm., or the total force in <i>CD</i> is 16×11 g., <i>i.e.</i>, each square +11 cm., or the total force in <i>CD</i> is 16×11 g., <i>i.e.</i>, each square centimeter of <i>CD</i> receives an additional force of 1 g. <i>Hence the force exerted on a unit area at</i> <i>AB</i> <i>is transmitted to every unit area within the vessel.</i></p></div> @@ -2544,7 +2502,7 @@ area that may be 100 or 1000 times that of the smaller. The smaller one is moved up and down by a lever; on each upstroke, liquid is drawn in from a reservoir, while each down-stroke forces some of the liquid into the -space about the large piston. Valves at <i>V</i> and <i>V´</i> prevent +space about the large piston. Valves at <i>V</i> and <i>V´</i> prevent the return of the liquid. If the area of <i>P</i> is 1,000 times that of <i>p</i>, then the force exerted by <i>P</i> is 1000 times the force employed in moving <i>p</i>. On the other hand, since @@ -2809,12 +2767,12 @@ Principle and the law of floating bodies.</p> depth of 3 ft. in the water. Find the weight of the boat. What weight of cargo will sink it to an average depth of 5 ft.?</p> -<p><b>Solution.</b>—The volume of the water displaced is 20 × 6 × 3 cu. -ft. = 360 cu. ft. Since 1 cu. ft. of water weighs 62.4 lbs., 360 × +<p><b>Solution.</b>—The volume of the water displaced is 20 × 6 × 3 cu. +ft. = 360 cu. ft. Since 1 cu. ft. of water weighs 62.4 lbs., 360 × 62.4 lbs. = 22,464 lbs., the weight of water displaced. By the law of floating bodies this is equal to the weight of the boat. When loaded -the volume of water displaced is 20 ft. × 6 × 5 ft. which equal 600 -cu. ft. 600 × 62.4 lbs. = 37,440 lbs. This is the weight of the water +the volume of water displaced is 20 ft. × 6 × 5 ft. which equal 600 +cu. ft. 600 × 62.4 lbs. = 37,440 lbs. This is the weight of the water displaced when loaded. 37,440 lbs. - 22,464 lbs. = 14,976 lbs., the weight of the cargo.</p> @@ -2910,7 +2868,7 @@ weighs 124.8 lbs.? If he weighs 187.2 lbs.?</p> volume? How much water does a floating block of wood displace if it weighs 125 lbs.? 125 g.? 2 kg.? 2000 kg.?</p> -<p>12. A flat boat 10 × 40 ft. in size will sink how much in the +<p>12. A flat boat 10 × 40 ft. in size will sink how much in the water when 10 horses each weighing 1250 lbs. are placed on board?</p> @@ -2983,7 +2941,7 @@ centimeter, therefore we have:</p> <p>In the English system, the density of water is 62.4 pounds per cubic foot, therefore in this system we have:</p> -<p>Density (lbs. per cu. ft.) = (numerically) 62.4 × sp. gr.</p> +<p>Density (lbs. per cu. ft.) = (numerically) 62.4 × sp. gr.</p> <p><span class="pagenum"><a name="Page_53" id="Page_53">[Pg 53]</a></span></p> @@ -3006,9 +2964,9 @@ density equals mass/volume; the specific gravity =</p> <p><b>(c) Solids Lighter than Water.</b>—This will require a sinker to hold the body under water. Weigh the solid in air (<i>w</i>). Weigh the sinker in water (<i>s</i>). Attach -the sinker to the solid and weigh both in water (w´). +the sinker to the solid and weigh both in water (w´). The specific gravity equals</p> -<div class="blockquot">(wt. of solid in air)/(loss in wt. of solid in water) or <i>w/((w + s) - w´)</i></div> +<div class="blockquot">(wt. of solid in air)/(loss in wt. of solid in water) or <i>w/((w + s) - w´)</i></div> <p>The apparent loss of weight of the solid is equal to the sum of its weight in air plus the weight of the sinker in water, @@ -3270,7 +3228,7 @@ of mercury at sea-level is 76 cm. Since the weight of 1 cc. of mercury is 13.6 grams, the pressure inside the tube at the level of the surface of the mercury in the dish is -equal to 1 × 76 × 13.6 or 1033.6 g. per +equal to 1 × 76 × 13.6 or 1033.6 g. per square centimeter. Therefore the <i>atmospheric pressure</i> on the surface of the mercury in the dish is 1033.6 g. per square centimeter, @@ -3364,9 +3322,9 @@ the level of the mercury surface in contact with the air. <p><span class="pagenum"><a name="Page_62" id="Page_62">[Pg 62]</a></span></p> -<p>3. What is the weight of the air in a room if it is 10 × 8 × 4 meters?</p> +<p>3. What is the weight of the air in a room if it is 10 × 8 × 4 meters?</p> -<p>4. What weight of air is in a room 10 × 15 × 10 ft.?</p> +<p>4. What weight of air is in a room 10 × 15 × 10 ft.?</p> <p>5. When smoke rises in a straight line from chimneys, is it an indication of a high or low barometric pressure? Why?</p> @@ -3466,10 +3424,10 @@ the volume to one-third and so on.</p> volume of a given mass of gas at constant temperature is inversely proportional to the pressure to which it is subjected</i>.</p> -<p>This law is often expressed mathematically. <i>P/P´ = V´/V</i>, or -<i>PV = P´V´</i>. Since doubling the pressure reduces the +<p>This law is often expressed mathematically. <i>P/P´ = V´/V</i>, or +<i>PV = P´V´</i>. Since doubling the pressure reduces the volume one-half, it doubles the density. Tripling the -pressure triples the density. We therefore have <i>P/P´ = D/D´</i> +pressure triples the density. We therefore have <i>P/P´ = D/D´</i> or the density of a gas directly proportional to its pressure.</p> <div class="figcenter" style="width: 400px;"> @@ -4058,7 +4016,7 @@ Illustrations and Applications:</p> <div class="blockquot"> -<p>Specific gravity, <i>W_{a}/(W_{a} - W_{w})</i>, <i>(W_{a} - W_{l})/(W_{a} - W_{w})</i>, Boyle's, <i>PV = P´V´</i></p></div> +<p>Specific gravity, <i>W_{a}/(W_{a} - W_{w})</i>, <i>(W_{a} - W_{l})/(W_{a} - W_{w})</i>, Boyle's, <i>PV = P´V´</i></p></div> <p>Devices: Hydraulic press, air cushion, barometer—mercurial and aneroid. Pumps, lift, force, vacuum, compression, centrifugal, @@ -4428,7 +4386,7 @@ has led to the <i>calculation</i> of what is called the "quantity of motion" of a body, or its <i>momentum</i>. It is computed by multiplying the mass by the velocity. If the C.G.S. system is used we shall have as the momentum of a 12 g. -body moving 25 cm. a second a momentum of 12 × 25 +body moving 25 cm. a second a momentum of 12 × 25 or 300 C.G.S. units of momentum. This unit has no name and is therefore expressed as indicated above. The formula for computing momentum is: <i>M = mv</i>.</p> @@ -4913,12 +4871,12 @@ hang horizontally when the board is suspended there. Weigh the board by a spring balance hung at <i>O</i>. This will be the resultant in the following tests. Now hang the board from two spring balances at <i>M</i> and <i>N</i> and read -both <i>balances</i>. Call readings <i>f</i> and <i>f´</i>. To test the forces +both <i>balances</i>. Call readings <i>f</i> and <i>f´</i>. To test the forces consider <i>M</i> as a fixed point (see Fig. 67) and the weight of the board to act at <i>O</i>. Then the moment of the weight of the board should be equal the moment of the force at <i>N</i> since the board does not move, or <i>w</i> times <i>OM</i> equals -<i>f´</i> times <i>NM</i>. If <i>N</i> is considered the fixed point then the +<i>f´</i> times <i>NM</i>. If <i>N</i> is considered the fixed point then the moment of the weight of the board and of <i>f with reference to the point N</i> should be equal, or <i>w</i> times <i>ON</i> = <i>f</i> times <i>NM</i>. Keeping this illustration in mind, the law of parallel forces @@ -4940,15 +4898,15 @@ the string is 5 ft. from the rear boy?</p> <p><b>Solution.</b>—The moment of the force <i>F</i> exerted about the opposite -end by the rear boy is <i>F</i> × 8. The +end by the rear boy is <i>F</i> × 8. The moment of the weight about the -same point is 40 × (8 - 5) = 120. -Therefore <i>F</i> × 8 = 120, or <i>F</i> = 15, +same point is 40 × (8 - 5) = 120. +Therefore <i>F</i> × 8 = 120, or <i>F</i> = 15, the force exerted by the rear boy. The front boy exerts a force of <i>F</i> -whose moment about the other end of the rod is <i>F</i> × 8. The -moment of the weight about the same point is 40 × 5 = 200. -Since the moment of <i>F</i> equals this, 200 = <i>F</i> × 8, or <i>F</i> = 25. Hence +whose moment about the other end of the rod is <i>F</i> × 8. The +moment of the weight about the same point is 40 × 5 = 200. +Since the moment of <i>F</i> equals this, 200 = <i>F</i> × 8, or <i>F</i> = 25. Hence the front boy exerts 25 lbs. and the rear boy 15 lbs.</p></div> <div class="figright" style="width: 200px;"> @@ -5400,7 +5358,7 @@ by "<i>g</i>" and the number of seconds by <i>t</i>, therefore the formula for finding the velocity, <i>V</i>,<a name="FNanchor_F_6" id="FNanchor_F_6"></a><a href="#Footnote_F_6" class="fnanchor">[F]</a> of a falling body starting from rest is <i>V</i> = <i>gt</i>. In studying gravity (Art. 89) we learned that its force varies as one moves toward -or away from the equator. (How?) In latitude 38° the +or away from the equator. (How?) In latitude 38° the acceleration of gravity is 980 cm. per second each second of time.</p> @@ -5616,7 +5574,7 @@ mass or the material of which the pendulum is made.</p> the length of the arc through which it swings.</p> <p>3. The period is directly proportional to the square root -of the length. Expressed mathematically, <i>t</i>/<i>t´</i> = √<i>l</i>/√<i>l´</i>.</p> +of the length. Expressed mathematically, <i>t</i>/<i>t´</i> = √<i>l</i>/√<i>l´</i>.</p> <p><b>103. Uses of the Pendulum.</b>—The chief use of the pendulum is to regulate motion in clocks. The wheels are @@ -5633,7 +5591,7 @@ the time in seconds of a single vibration and <i>l</i> the length of the pendulum. If, for example, the length of the seconds pendulum is 99.31 cm., then 1 = π√(99.31/<i>g</i>); squaring both sides of the equation, we have 1<sup>2</sup> = π<sup>2</sup>(99.31/<i>g</i>), or <i>g</i> = -<span class="pagenum"><a name="Page_117" id="Page_117">[Pg 117]</a></span>π<sup>2</sup> × 99.31/1<sup>2</sup> = 980.1 cm. per sec., per sec. From this it follows +<span class="pagenum"><a name="Page_117" id="Page_117">[Pg 117]</a></span>π<sup>2</sup> × 99.31/1<sup>2</sup> = 980.1 cm. per sec., per sec. From this it follows that, since the force of gravity depends upon the distance from the center of the earth, the pendulum may be used to determine the elevation of a place above sea @@ -5777,7 +5735,7 @@ and distance units employed. One, the i, is equal to lbs. force, how much work is done?</p> <p><b>Solution.</b>—Since the force employed is 500 lbs., and the distance -is 2 × 5280 ft., the work done is 500 × 2 × 5280 or 5,280,000 +is 2 × 5280 ft., the work done is 500 × 2 × 5280 or 5,280,000 ft.-lbs.</p></div> <p><b>106. Energy.</b>—In the various cases suggested in the @@ -5815,7 +5773,7 @@ just mentioned should fall from its elevated position upon a post, it could drive the post into the ground because its motion at the instant of striking enables it to do work. To compute potential energy you compute the work done upon -the body. That is, <i>P.E.</i> = <i>w</i> × <i>h</i> or <i>f</i> × <i>s</i>.</p> +the body. That is, <i>P.E.</i> = <i>w</i> × <i>h</i> or <i>f</i> × <i>s</i>.</p> <p><b>108. Kinetic Energy.</b>—<i>The energy due to the motion of a body is called kinetic energy</i>. The amount of kinetic @@ -5837,18 +5795,18 @@ as follows, in solving the above problem:</p> <p>First, find the velocity of the falling body which has fallen 16 ft. A body falls 16 ft. in <i>one</i> second. In this time it gains a velocity of 32 ft. per second. Now using the formula for kinetic energy -<i>K.E.</i> = <i>wv</i><sup>2</sup>/(2<i>g</i>), we have <i>K.E.</i> = 100 × 32 × 32/(2 × 32) = 1600 ft.-lbs. as before. +<i>K.E.</i> = <i>wv</i><sup>2</sup>/(2<i>g</i>), we have <i>K.E.</i> = 100 × 32 × 32/(2 × 32) = 1600 ft.-lbs. as before. The formula, <i>K.E.</i> = <i>wv</i><sup>2</sup>/(2<i>g</i>), may be derived in the following manner:</p> <p>The kinetic energy of a falling body equals the work done in giving -it its motion, that is, <i>K.E.</i> = <i>w</i> × <i>S</i>, in which, <i>w</i> = the weight +it its motion, that is, <i>K.E.</i> = <i>w</i> × <i>S</i>, in which, <i>w</i> = the weight of the body and <i>S</i> = the distance the body must fall freely in order <span class="pagenum"><a name="Page_122" id="Page_122">[Pg 122]</a></span>to acquire its velocity. The distance fallen by a freely falling body, <i>S</i>, = 1/2<i>gt</i><sup>2</sup> = <i>g</i><sup>2</sup><i>t</i><sup>2</sup>/(2<i>g</i>) (Art. 98, p. 111). Now, <i>v</i> = <i>gt</i> and <i>v</i><sup>2</sup> = <i>g</i><sup>2</sup><i>t</i><sup>2</sup>.</p> <p>Substituting for <i>g</i><sup>2</sup><i>t</i><sup>2</sup>, its equal <i>v</i><sup>2</sup>, we have <i>S</i> = <i>v</i><sup>2</sup>/(2<i>g</i>). Substituting -this value of S in the equation <i>K.E.</i> = <i>w</i> × <i>S</i>, we have <i>K.E.</i> +this value of S in the equation <i>K.E.</i> = <i>w</i> × <i>S</i>, we have <i>K.E.</i> = <i>wv</i><sup>2</sup>/(2<i>g</i>).</p> <p>Since the kinetic energy of a moving body depends upon its mass @@ -6124,7 +6082,7 @@ work have they the same power? Explain.</p> <p>9. If 160 cu. ft. of water flow each second over a dam 15ft. high what is the available power?</p> -<p>10. What power must an engine have to fill a tank 11 × 8 × 5 ft. +<p>10. What power must an engine have to fill a tank 11 × 8 × 5 ft. with water 120 ft. above the supply, in 5 minutes?</p> <p>11. A hod carrier weighing 150 lbs. carries a load of bricks weighing @@ -6278,7 +6236,7 @@ equals the resistance overcome.</p> <p>If <i>f</i> represents the force or effort, and <i>D<sub>f</sub></i> the space it acts through, and <i>w</i> represents the weight or resistance, and <i>D<sub>w</sub></i> the space it acts through, then the law of machines -may be expressed by an equation, <i>f × D<sub>f</sub> = w × D<sub>w</sub></i>. +may be expressed by an equation, <i>f × D<sub>f</sub> = w × D<sub>w</sub></i>. That is, <i>the effort times the distance the effort acts equals the resistance times the distance the resistance is moved or overcome</i>. When the product of two numbers equals the @@ -6349,7 +6307,7 @@ it to the axis as well as upon the magnitude of the force.</p> <div class="figleft" style="width: 250px;"> <img src="images/i_152.png" width="250" height="154" alt="" /> <span class="caption"><span class="smcap">Fig.</span> 87.—The two moments are equal -about <i>C</i>. 100 × 15 = 50 × 30</span> +about <i>C</i>. 100 × 15 = 50 × 30</span> </div> <p>From the experiment @@ -6357,8 +6315,8 @@ just described, the moment of the acting force equals the moment of the -weight or <i>f × D<sub>f</sub> = -w × D<sub>w</sub></i>, or the effort +weight or <i>f × D<sub>f</sub> = +w × D<sub>w</sub></i>, or the effort times the effort arm equals the weight times the weight arm. This equation is called the law of the lever. It corresponds to the general @@ -6375,7 +6333,7 @@ finding the ratio of the resistance or weight to the effort. What must be the relative lengths of the effort arm and resistance or weight arm in the example just mentioned? Since the effort times the effort arm equals the weight -times the weight arm, if <i>f × D<sub>f</sub> = w × D<sub>w</sub></i>, then <i>D<sub>f</sub></i> is +times the weight arm, if <i>f × D<sub>f</sub> = w × D<sub>w</sub></i>, then <i>D<sub>f</sub></i> is five times <i>D<sub>w</sub></i>. Hence the mechanical advantage of a lever is easily found by finding the ratio of the effort arm to the weight arm.</p> @@ -6524,7 +6482,7 @@ and the resisting weight <i>W</i> at the extremity of a radius of the axle. Hence, if <i>D<sub>f</sub></i>, the effort distance, is three times <i>D<sub>w</sub></i>, the weight distance, the weight that can be supported is three times the effort. Here as in the lever,<span class="pagenum"><a name="Page_139" id="Page_139">[Pg 139]</a></span> -<i>f × D<sub>f</sub> = w × D<sub>w</sub></i>, or <i>w:f = D<sub>f</sub>:D<sub>w</sub></i>, or <i>the ratio of the +<i>f × D<sub>f</sub> = w × D<sub>w</sub></i>, or <i>w:f = D<sub>f</sub>:D<sub>w</sub></i>, or <i>the ratio of the weight to the effort equals the ratio of the radius of the wheel to the radius of the axle</i>. This is therefore the mechanical advantage of the wheel and axle. Since the diameters @@ -6684,7 +6642,7 @@ machine. If there were no friction or wasted work, the efficiency would be perfect, or, as it is usually expressed, would be 100 per cent. Consider a system of pulleys into which are put 600 ft.-lbs. of work. With 450 ft.-lbs. of -useful work resulting, the efficiency would be 450 ÷ +useful work resulting, the efficiency would be 450 ÷ 600 = {3/4}, or 75 per cent. In this case 25 per cent. of the<span class="pagenum"><a name="Page_143" id="Page_143">[Pg 143]</a></span> work done on the machine is wasted. In a simple lever the friction is slight so that nearly 100 per cent. efficiency @@ -6767,10 +6725,10 @@ the lever, or at the circumference of the wheel. While the effort moves once about the circumference of the wheel the weight is pushed forward a distance equal to the distance between two threads (the pitch of the screw). The work -done by the effort therefore equals <i>F × 2πr</i>, <i>r</i> being the +done by the effort therefore equals <i>F × 2πr</i>, <i>r</i> being the radius of the wheel, and the work done on the weight -equals <i>W × s</i>, <i>s</i> being the pitch of the screw. By the law -of machines <i>F × 2πr = W × s</i> or <i>W / F +equals <i>W × s</i>, <i>s</i> being the pitch of the screw. By the law +of machines <i>F × 2πr = W × s</i> or <i>W / F = (2πr) / s</i>. Therefore the mechanical advantage of the screw equals <i>(2πr) / s</i>. Since the distance the weight moves is @@ -7044,7 +7002,7 @@ weighing 200 lbs. is just able to lift 600 lbs. What is the efficiency of the system?</p> <p>19. What is the horse-power of a pump that can pump out a cellar -full of water 40 ft. × 20 ft. by 10 ft. deep, in 30 minutes?</p> +full of water 40 ft. × 20 ft. by 10 ft. deep, in 30 minutes?</p> <p>20. How many tons of coal can a 5 horse-power hoisting engine raise in 30 minutes from a barge to the coal pockets, a height @@ -7225,14 +7183,14 @@ What is the efficiency?</p> <p>Work; how measured, units, foot-pound, kilogram meter, erg.</p> -<p>Energy; how measured, units, potential, <i>P.E.</i> = <i>w × h</i>, or <i>f × s</i>. +<p>Energy; how measured, units, potential, <i>P.E.</i> = <i>w × h</i>, or <i>f × s</i>. Kinetic = <i>(wv<sup>2</sup>)/(2g)</i>.</p> <p>Power; how measured, units, horse power, watt, 5 forms of energy, -conservation. H.p. = (lbs. × ft.)/(550 × sec.).</p> +conservation. H.p. = (lbs. × ft.)/(550 × sec.).</p> -<p>Machines; 6 simple forms, 2 groups, advantages, uses, Law: <i>W × D<sub>w</sub></i> -= <i>F × D<sub>f</sub></i>.</p> +<p>Machines; 6 simple forms, 2 groups, advantages, uses, Law: <i>W × D<sub>w</sub></i> += <i>F × D<sub>f</sub></i>.</p> <p>Lever; moments, mechanical advantage, uses and applications.</p> @@ -7438,10 +7396,10 @@ in this country are the <i>Centigrade</i> and the <i>Fahrenheit</i>. The <i>Fahrenheit thermometer scale</i> has the temperature of melting ice marked -32°. The boiling point or steam +32°. The boiling point or steam temperature of pure water under standard conditions of atmospheric -pressure is marked 212° +pressure is marked 212° and the space between these two fixed points is divided into 180 parts.</p> @@ -7466,10 +7424,10 @@ necessary to express in centigrade degrees a temperature for which the Fahrenheit reading is given or <i>vice versa</i>. Since there are 180 Fahrenheit degrees between the "fixed points" and 100 centigrade degrees, the Fahrenheit -degrees are smaller than the centigrade, or 1°F. = -5/9°C. and 1°C. = 9/5°F. One must also take into account +degrees are smaller than the centigrade, or 1°F. = +5/9°C. and 1°C. = 9/5°F. One must also take into account the fact that the melting point of ice on the Fahrenheit -scale is marked 32°. Hence the following rule: To +scale is marked 32°. Hence the following rule: To change a Fahrenheit reading to centigrade subtract 32<span class="pagenum"><a name="Page_164" id="Page_164">[Pg 164]</a></span> and take 5/9 of the remainder, while to change centigrade to Fahrenheit multiply the centigrade by 9/5 and add 32 @@ -7478,12 +7436,12 @@ following formulas.</p> <div class="blockquot"> -<p>(F.° - 32)5/9 = C.°, 9C.°/5 + 32° = F.°</p></div> +<p>(F.° - 32)5/9 = C.°, 9C.°/5 + 32° = F.°</p></div> <p>Another method of changing from one thermometric scale to another is as follows:</p> -<p>A temperature of -40°F. is also <i>represented</i> by -40°C., therefore +<p>A temperature of -40°F. is also <i>represented</i> by -40°C., therefore to change a Fahrenheit reading into centigrade, we add 40 to the given reading, then divide by 1.8 after which subtract 40. To change from a centigrade to Fahrenheit reading the only difference @@ -7517,46 +7475,46 @@ point, the ice, like any other solid, cools to the temperature of the surrounding space. For example, a piece of ice out of doors -is at 10°F. when the air is at this temperature. It follows +is at 10°F. when the air is at this temperature. It follows then, that when ice has been cooled below the freezing<span class="pagenum"><a name="Page_165" id="Page_165">[Pg 165]</a></span> temperature that heat is required to warm the ice up to its melting point; or in other words that ice at its melting temperature possesses some heat. The temperature at which absolutely no heat exists is called <i>absolute zero</i>. There has been devised an <i>absolute scale of</i> temperature. This -scale is based upon the centigrade scale, <i>i.e.</i>, with 100° +scale is based upon the centigrade scale, <i>i.e.</i>, with 100° between the two fixed points; the scale, however, extends -down, below the centigrade zero, 273°, to what is called +down, below the centigrade zero, 273°, to what is called <i>absolute zero</i>. It follows therefore that upon the absolute scale, the melting point of ice, and the boiling point of -water are 273° and 373° respectively. (See Fig. 124.)</p> +water are 273° and 373° respectively. (See Fig. 124.)</p> <p>The means employed to find the location of absolute zero are of much interest. It has been observed that when heated a gas tends to expand. If a measured volume -of air at 0°C. is cooled or heated 1°C., it changes its +of air at 0°C. is cooled or heated 1°C., it changes its volume 1/273, the pressure remaining the same. If it is -cooled 10° it loses 10/273, if cooled 100° it loses 100/273 and +cooled 10° it loses 10/273, if cooled 100° it loses 100/273 and so on. No matter how far it is cooled the same rate of reduction continues as long as it remains in the gaseous state. From these facts it is concluded that if the cooling -could be carried down 273° that the volume would be reduced +could be carried down 273° that the volume would be reduced 273/273 or that the volume of the gas would be reduced to nothing. This is believed to mean that the molecular motion constituting heat would cease rather than that the matter composing the gas would disappear. Scientists have been able to obtain temperatures of extreme cold far down on the absolute scale. Liquid air -has a temperature of -292°F., or -180°C. or 93°A. -The lowest temperature thus far reported is 1.7°A. -or -271.3°C., obtained in 1911, by evaporating liquid +has a temperature of -292°F., or -180°C. or 93°A. +The lowest temperature thus far reported is 1.7°A. +or -271.3°C., obtained in 1911, by evaporating liquid helium.</p> <p><b>145. The Law of Charles.</b>—The facts given in the last -paragraph mean that if 273 ccm. of a gas at 0°C. or 273° -A. are cooled 100°, or to -100°C., or 173°A., then it<span class="pagenum"><a name="Page_166" id="Page_166">[Pg 166]</a></span> +paragraph mean that if 273 ccm. of a gas at 0°C. or 273° +A. are cooled 100°, or to -100°C., or 173°A., then it<span class="pagenum"><a name="Page_166" id="Page_166">[Pg 166]</a></span> will lose 100/273 of its volume or have a volume of 173 -ccm. If warmed 100°, or up to 100°C., or 373°A., it +ccm. If warmed 100°, or up to 100°C., or 373°A., it will have a volume of 373 ccm. It follows then that in every case the volume will correspond to its absolute temperature, providing the pressure remains unchanged. @@ -7573,7 +7531,7 @@ sometimes combined into one expression as follows:</p> <div class="blockquot"> -<p><i>PV/T = P´V´/T´</i></p></div> +<p><i>PV/T = P´V´/T´</i></p></div> <p>or the product of the volume and pressure of a constant mass of gas is proportional to its absolute temperature.</p> @@ -7597,25 +7555,25 @@ and Charles.</p> Express the temperature of freezing water on the three thermometer scales.</p> -<p>2. A comfortable room temperature is 68°F. What is this temperature +<p>2. A comfortable room temperature is 68°F. What is this temperature on the centigrade and absolute scales?</p> -<p>3. Change a temperature of 15°C. to F.; 15°F. to C.; -4°C. to F.; --20°F. to C.</p> +<p>3. Change a temperature of 15°C. to F.; 15°F. to C.; -4°C. to F.; +-20°F. to C.</p> <p><span class="pagenum"><a name="Page_167" id="Page_167">[Pg 167]</a></span></p> -<p>4. The temperature of the human body is 98.6°F. What is this +<p>4. The temperature of the human body is 98.6°F. What is this temperature on the absolute and centigrade scales?</p> -<p>5. The temperature of liquid air is -180°C. What is it on the +<p>5. The temperature of liquid air is -180°C. What is it on the Fahrenheit scale?</p> -<p>6. Mercury is a solid at -40°F. What is this on the centigrade +<p>6. Mercury is a solid at -40°F. What is this on the centigrade scale?</p> <p>7. How much heat will be required to raise the temperature of -8 lbs. of water 32°F.; 5 lbs. 10°F.?</p> +8 lbs. of water 32°F.; 5 lbs. 10°F.?</p> <div class="figcenter" style="width: 400px;"> <img src="images/i_185.png" width="400" height="61" alt="" /> @@ -7624,7 +7582,7 @@ body.</span> </div> <p>8. How much heat will be required to raise the temperature of -30 g. of water 43°C.; 20 g., 50°C.?</p> +30 g. of water 43°C.; 20 g., 50°C.?</p> <p>9. Compute the temperature of absolute zero on the Fahrenheit scale.</p> @@ -7635,7 +7593,7 @@ the cold and after a few minutes both are placed in the lukewarm water, this water will feel cool to one hand and warm to the other. Explain.</p> -<p>11. If 200 ccm. of air at 200° absolute is heated to 300°A. under +<p>11. If 200 ccm. of air at 200° absolute is heated to 300°A. under constant pressure, what volume will the air occupy at the latter temperature?</p> @@ -7657,8 +7615,8 @@ bulb blown on a glass tube and inverted in a dish of water. (See Fig. 1.) The <i>water thermometer</i> consists of a glass bulb filled with water which rises into a tube attached to the bulb. One disadvantage of the water thermometer -is its limited range since it cannot be used below 0° or -above 100°. Why?</p> +is its limited range since it cannot be used below 0° or +above 100°. Why?</p> <p><b>147. Expansion of Liquids.</b>—The expansion of liquids differs from that of gases @@ -7669,7 +7627,7 @@ than gases. The <i>rate</i> of expansion per degree is called the <i>Coefficient of Expansion</i>. For example, the coefficient of expansion of a gas under constant pressure -at 0°C. is {1/273} of its volume per degree +at 0°C. is {1/273} of its volume per degree centigrade.</p> <p>(b) Different liquids expand at wholly @@ -7688,9 +7646,9 @@ thermometer.</span> <p>(c) The same liquid often has different coefficients of expansion at different temperatures. Water between -5°C. and 6°C. has a coefficient expansion of 0.00002 per -degree centigrade, between 8° and 50° of 0.0006, between -99° and 100° of 0.00076. The coefficient of expansion of +5°C. and 6°C. has a coefficient expansion of 0.00002 per +degree centigrade, between 8° and 50° of 0.0006, between +99° and 100° of 0.00076. The coefficient of expansion of mercury, however, is constant for a wide range of temperature and, therefore, it is well adapted for use in thermometers.</p> @@ -7713,7 +7671,7 @@ after the water cools to a certain temperature that <i>expansion of the water occurs with further cooling</i>.</p></div> <p>Careful tests show that the water on cooling -contracts until it reaches 4°C. On cooling +contracts until it reaches 4°C. On cooling below this temperature it expands. For this reason, when the water of a lake or river freezes, the coldest water is at the surface. On @@ -7728,8 +7686,8 @@ the ice melt. The result would be that fish and other aquatic life would be killed. Climate would be so changed that the earth might become uninhabitable. Since water is densest -at 4°C. all the water in a lake or river, when -it is covered with ice, is at 4°C. except that +at 4°C. all the water in a lake or river, when +it is covered with ice, is at 4°C. except that near the surface.</p> <div class="figright" style="width: 80px;"> @@ -7827,9 +7785,9 @@ water and its results.</p> <h4>Exercises</h4> <p>1. The gas within a partly inflated balloon has a volume of 1000 -cu. ft. at a pressure of 74 cm., and a temperature of 15°C. +cu. ft. at a pressure of 74 cm., and a temperature of 15°C. What will be the volume of the gas when its pressure is 37 cm. -and the temperature is -17°C.?</p> +and the temperature is -17°C.?</p> <p>2. A man taking a full breath on the top of a mountain fourteen thousand feet high inhales 4 liters of air, the pressure being 40 @@ -7839,58 +7797,58 @@ and the temperature is the same as on the mountain top?</p> <p>3. If the coefficient of linear expansion of iron is 0.000012 per<span class="pagenum"><a name="Page_172" id="Page_172">[Pg 172]</a></span> degree C., how much will an iron bridge 1000 ft. long change -in length in warming from -20°C. on a winter day to 30°C. +in length in warming from -20°C. on a winter day to 30°C. upon a summer day.</p> <p>4. What are some of the results that would follow in freezing weather if water continually contracted on being cooled to -zero instead of beginning to expand when cooled below 4°C.?</p> +zero instead of beginning to expand when cooled below 4°C.?</p> <p>5. Mention two instances that you have noticed of expansion occurring when a body is heated?</p> -<p>6. Compare the density of air at 30°C. with that at 10°C. at the +<p>6. Compare the density of air at 30°C. with that at 10°C. at the same pressure. If both are present in a room, where will each be found? Why?</p> -<p>7. Compare the density of water at 40°C. with that at 10°C. If +<p>7. Compare the density of water at 40°C. with that at 10°C. If water at the two temperatures are in a tank, where will each be found? Why?</p> -<p>8. If water at 0°C. and at 4°C. are both in a tank, where will +<p>8. If water at 0°C. and at 4°C. are both in a tank, where will each be found? Why?</p> <p>9. How much heat will be required to raise the temperature of a -cubic foot of water 10°F.?</p> +cubic foot of water 10°F.?</p> <p>10. How much heat will be required to raise the temperature of -4 liters of water 25°C.?</p> +4 liters of water 25°C.?</p> <p>11. How much longer would the cables of the Brooklyn suspension -bridge be on a summer's day when the temperature is 30°C. -than in winter at -20°C., the length of cable between the supports +bridge be on a summer's day when the temperature is 30°C. +than in winter at -20°C., the length of cable between the supports being about 1600 ft.</p> -<p>12. If 25 liters of air at -23°C. is warmed to 77°C. under constant +<p>12. If 25 liters of air at -23°C. is warmed to 77°C. under constant pressure, what will be the resulting volume of air? Explain.</p> -<p>13. White pig iron melts at about 2000°F. Express this temperature +<p>13. White pig iron melts at about 2000°F. Express this temperature upon the centigrade and absolute scales.</p> -<p>14. If 200 ccm. of air at 76 cm. pressure and 27°C. temperature be -heated to 127°C. at a pressure of 38 cm. what will be the resulting +<p>14. If 200 ccm. of air at 76 cm. pressure and 27°C. temperature be +heated to 127°C. at a pressure of 38 cm. what will be the resulting volume?</p> <p>15. A balloon contains 10,000 cu. ft. of gas at 75.2 cm. pressure and -24°C. It ascends until the pressure is 18 cm. and the temperature -is -10°C. What is the volume of gas it then contains.</p> +24°C. It ascends until the pressure is 18 cm. and the temperature +is -10°C. What is the volume of gas it then contains.</p> <p>16. A gas holder contains 50 "cu. ft." of gas at a pressure of one -atmosphere and 62°F. How much gas will it hold at 10 atmospheres -and 32°F.</p> +atmosphere and 62°F. How much gas will it hold at 10 atmospheres +and 32°F.</p> <p>17. One thousand "cubic feet" of illuminating gas has what volume -with 75 lbs. pressure and temperature of 10°C.</p> +with 75 lbs. pressure and temperature of 10°C.</p> <p>18. Define a "cubic foot" of illuminating gas.</p> @@ -8138,7 +8096,7 @@ earth's surface perpendicular to the sun's rays were made at Mt. Wilson in 1913. The average of 690 observations gave a value of 1.933 calories per minute. These results indicate that the sun's radiation per square centimeter is -sufficient to warm 1 g. of water 1.933°C. each minute. +sufficient to warm 1 g. of water 1.933°C. each minute. Although the <i>nature</i> of <i>radiation</i> is not discussed until Art. 408-411 in light, it should be said here that all bodies are radiating heat waves at all temperatures, the heat @@ -8205,7 +8163,7 @@ applied at the <i>top</i> of a test-tube of water, the hot water being lighter is found at the top, while at the bottom the water remains cold. On the other hand, if heat is applied at the <i>bottom</i> of the vessel, as soon as the water at the bottom -is warmed (above 4°C.) it expands, becomes lighter +is warmed (above 4°C.) it expands, becomes lighter and is pushed up to the top by the colder, denser water about it. This circulation of water continues as long as heat @@ -8451,8 +8409,8 @@ radiators the pressure of the air closes the valve, making a partial vacuum<span class="pagenum"><a name="Page_188" id="Page_188">[Pg 188]</a></span> inside. The boiling point of water falls as the pressure upon it is reduced. As water will not boil under ordinary -atmospheric pressure until its temperature is 100°C. -(212°F.), it follows that by the use of vacuum systems, +atmospheric pressure until its temperature is 100°C. +(212°F.), it follows that by the use of vacuum systems, often called vapor systems, of steam heating, water will be giving off hot vapor even after the fire has been banked for hours. This results in a considerable saving of fuel.</p> @@ -8465,7 +8423,7 @@ hours. This results in a considerable saving of fuel.</p> <p><b>164. The Plenum System of Heating.</b>—In the plenum system of heating (see Fig. 147) fresh air is drawn through a window from outdoors and goes first through tempering -coils where the temperature is raised to about 70°. The +coils where the temperature is raised to about 70°. The fan then forces some of the air through heating coils, where it is reheated and raised to a much higher temperature, depending upon the weather conditions. Both @@ -8564,7 +8522,7 @@ be the direction of the air currents at the top and at the bottom of the door? Explain.</p> <p>8. If a hot-water heating system contains 100 cu. ft. of water -how much heat will be required to raise its temperature 150°F.?</p> +how much heat will be required to raise its temperature 150°F.?</p> <p>9. Why does a tall chimney give a better draft than a short one?</p> @@ -8683,8 +8641,8 @@ temperature</i>.</p> <div class="blockquot"> -<p>For example, if the dew point is 5°C. and the temperature of -the air is 22°C., we find the densities of the water vapor at the two +<p>For example, if the dew point is 5°C. and the temperature of +the air is 22°C., we find the densities of the water vapor at the two temperatures, and find their ratio: 6.8/19.3 = 35 per cent. nearly. Determinations of humidity may give indication of rain or frost and are regularly made at weather bureau stations. They are also @@ -8693,12 +8651,12 @@ to see if the air is moist enough. For the most healthful conditions the relative humidity should be from 40 per cent. to 50 per cent.</p></div> <p><span class="smcap">Weight of Water</span> (<i>w</i>) <span class="smcap">in Grams Contained in 1 Cubic Meter of -Saturated Air at Various Temperatures</span> (<i>t</i>°)C.</p> +Saturated Air at Various Temperatures</span> (<i>t</i>°)C.</p> <div class="center"> <table border="0" cellpadding="4" cellspacing="0" summary=""> -<tr><td align="right"><i>t</i>°C.</td><td align="left"> <i>w</i></td></tr> +<tr><td align="right"><i>t</i>°C.</td><td align="left"> <i>w</i></td></tr> <tr><td align="right">-10 </td><td align="left"> 2.1</td></tr> <tr><td align="right">- 9 </td><td align="left"> 2.4</td></tr> <tr><td align="right">- 8 </td><td align="left"> 2.7</td></tr> @@ -8812,8 +8770,8 @@ ocean.</p> <p>5. Look up the derivation of the term "hygrometer." Give the use of the instrument.</p> -<p>6. Find the relative humidity of air at 20°C. if its dew point is -at 10°C.</p> +<p>6. Find the relative humidity of air at 20°C. if its dew point is +at 10°C.</p> <p>7. How may the relative humidity of the air in a home be increased?</p> @@ -8875,8 +8833,8 @@ that often the drop of water is frozen. The lowest temperatures are obtained by evaporating liquids at reduced pressure.</p></div> <p>Onnes by evaporating liquid helium at a pressure of -about 1.2 mm. reached the lowest temperature yet attained, -456°F., -or -271.3°C.</p> +about 1.2 mm. reached the lowest temperature yet attained, -456°F., +or -271.3°C.</p> <div class="blockquot"> @@ -9012,7 +8970,7 @@ ratio of the amount of heat required to change the temperature of a given mass of a substance 1 C. degree to the amount of heat required to change the temperature of the same mass of water 1 C. degree.</i> By definition, it requires 1 calorie to -raise the temperature of the gram of water 1°C. The +raise the temperature of the gram of water 1°C. The <i>specific heat</i> therefore of water is taken as one. The specific heat of most substances except hydrogen, is <i>less</i> than that of water, and as a rule, the denser the body the @@ -9047,13 +9005,13 @@ called the <i>method of mixtures</i>.</p> <p>For example, a definite weight of a substance, say a 200-g. iron ball, is placed in boiling water until it has the temperature of the -hot water, 100°C. Suppose that 300 g. of water at 18°C. be placed +hot water, 100°C. Suppose that 300 g. of water at 18°C. be placed in a calorimeter, and that the hot iron ball on being placed in the -water raises its temperature to 23.5°C. The heat received by the -water equals 5.5 × 300 = 1650 calories. This must have come -from the heated iron ball. 200 g. of iron then in cooling 76.5°C. -(100°-23.5°) gave out 1650 calories. Then 1 g. of iron in cooling -76.5°C. Would give out 8.25 calories or 1 g. of iron cooling 1°C. +water raises its temperature to 23.5°C. The heat received by the +water equals 5.5 × 300 = 1650 calories. This must have come +from the heated iron ball. 200 g. of iron then in cooling 76.5°C. +(100°-23.5°) gave out 1650 calories. Then 1 g. of iron in cooling +76.5°C. Would give out 8.25 calories or 1 g. of iron cooling 1°C. would yield about 0.11 calorie. The specific heat of the iron is then 0.11. For accurate determination the heat received by the calorimeter must be considered.</p></div> @@ -9074,20 +9032,20 @@ hot-water bottles and hot-water heating effective.</p> <div class="blockquot"> -<p>If one takes a pound of ice at 0°C. in one dish and a pound of -water at 0°C. in another, and warms the dish of ice by a Bunsen +<p>If one takes a pound of ice at 0°C. in one dish and a pound of +water at 0°C. in another, and warms the dish of ice by a Bunsen flame until the ice is just melted, and then warms the water in the other dish for the same time, the water will be found to be <i>hot</i> and -at a temperature 80°C., or 176°F.</p></div> +at a temperature 80°C., or 176°F.</p></div> <p><b>180. The Heat of Fusion of Ice.</b>—This experiment indicates the large amount of heat required to change the ice to water without changing its temperature. As indicated<span class="pagenum"><a name="Page_202" id="Page_202">[Pg 202]</a></span> by the experiment, it requires 80 calories to melt 1 g. of ice without changing its temperature or, in other words, -if one placed 1 g. of ice at 0°C. in 1 g. of water at 80°C., +if one placed 1 g. of ice at 0°C. in 1 g. of water at 80°C., the ice would be melted and the water would be cooled to -0°C.</p> +0°C.</p> <p><b>181. Heat Given out by Freezing water.</b>—Just as 80 calories of heat are required to melt 1 g. of ice, so in freezing @@ -9123,12 +9081,12 @@ a solid to a liquid without a change of temperature.</p> <p><b>182. Melting of Crystalline and Amorphous Substances.</b>—If a piece of ice is placed in boiling hot water and then removed, the temperature of the unmelted ice is still -0°C. There is no known means of warming ice under<span class="pagenum"><a name="Page_203" id="Page_203">[Pg 203]</a></span> +0°C. There is no known means of warming ice under<span class="pagenum"><a name="Page_203" id="Page_203">[Pg 203]</a></span> atmospheric pressure above its melting point and maintaining its solid state. Ice being composed of ice crystals is called a crystalline body. All crystalline substances have fixed melting points. For example, ice always melts at -0°C. The melting points of some common crystalline +0°C. The melting points of some common crystalline substances are given below:</p> <p><i>Melting Points of Some Crystalline Substances</i></p> @@ -9197,7 +9155,7 @@ of rocks.</p> </div> <p>Since water expands on freezing, pressure would on compressing -ice at 0°C., tend to turn it into water. Pressure +ice at 0°C., tend to turn it into water. Pressure does lower the melting point of ice, so that a little ice may melt when it is subjected to pressure. On removing the pressure the water freezes. This may be shown by placing @@ -9225,25 +9183,25 @@ melted ice refreezing above the wire.</p> <p>2. What are two advantages in the expansion of water while freezing?</p> <p>3. How much heat will be required to melt 1000 g. of ice and warm -the water to 20°C.?</p> +the water to 20°C.?</p> -<p>4. How many grams of ice at 0°C. can be melted by 400 g. of water -at 55°C.?</p> +<p>4. How many grams of ice at 0°C. can be melted by 400 g. of water +at 55°C.?</p> <p>5. What are two advantages of the high specific heat of water? Two disadvantages?</p> -<p>6. If the specific heat of iron is 0.1125, how much ice at 0°C. can -be melted by a 200-g. ball of iron heated to 300°C?</p> +<p>6. If the specific heat of iron is 0.1125, how much ice at 0°C. can +be melted by a 200-g. ball of iron heated to 300°C?</p> <p>7. What is the temperature of a hot ball of iron weighing 80 g., -if when placed on a piece of ice at 0°C. it melts 90 g. of ice?</p> +if when placed on a piece of ice at 0°C. it melts 90 g. of ice?</p> -<p>8. If 500 g. of copper at 400°C. are placed into 3000 g. of water -at 10°C. what will be the resulting temperature?</p> +<p>8. If 500 g. of copper at 400°C. are placed into 3000 g. of water +at 10°C. what will be the resulting temperature?</p> -<p>9. What weight of water at 90°C. will just melt 10 kg. of ice at -0°C.?</p> +<p>9. What weight of water at 90°C. will just melt 10 kg. of ice at +0°C.?</p> <p>10. If the smooth dry surface of two pieces of ice are pressed together for a short time the two pieces will be frozen into one @@ -9322,9 +9280,9 @@ or snow, the vapor within is condensed and the pressure upon the water is reduced. Vigorous boiling begins at once. By condensing the vapor repeatedly the water may be made to boil at the room temperature. At the top of -Mt. Blanc water boils at 84°C. While in steam boilers at +Mt. Blanc water boils at 84°C. While in steam boilers at 225 lbs. pressure to the square inch the boiling point is -nearly 200°C.</p> +nearly 200°C.</p> <p><b>186. Laws of Boiling.</b>—The following statements have been found by experiments to be true.</p> @@ -9357,7 +9315,7 @@ solid substances in a liquid raises its boiling point, additional energy being needed to overcome the adhesion involved in the solution. The boiling point is also affected by the character of the vessel containing the liquid. In -glass the boiling point is 101°.</p> +glass the boiling point is 101°.</p> <div class="figright" style="width: 300px;"> <img src="images/i_227.jpg" width="300" height="395" alt="" /> @@ -9427,7 +9385,7 @@ liquid attract the molecules of the solid and thus assist the change of state.</p> <p><b>189. Freezing Mixtures.</b>—If one attempts to freeze a -solution of salt and water, ice will not form at 0°C. but +solution of salt and water, ice will not form at 0°C. but several degrees lower. The ice formed however is pure. Evidently the attraction of the molecules of salt for the water molecules prevented the formation of ice until the @@ -9435,7 +9393,7 @@ motions of the water molecules had been reduced more than is necessary in pure water. As the temperature of freezing water is that of melting ice, ice in a salt solution melts at lower temperature than in pure water. In a -saturated salt solution this temperature is -22°C. It<span class="pagenum"><a name="Page_211" id="Page_211">[Pg 211]</a></span> +saturated salt solution this temperature is -22°C. It<span class="pagenum"><a name="Page_211" id="Page_211">[Pg 211]</a></span> is for this reason that the mixture of ice and salt used in freezing cream is so effective, the salt water in melting the ice, being cooled to a temperature many degrees below @@ -9448,7 +9406,7 @@ also produced by permitting the rapid evaporation of liquids under pressure. Carbon dioxide under high pressure is a liquid, but when allowed to escape into the air evaporates so rapidly that a portion of the liquid is frozen -into solid carbon dioxide which has a temperature of -80° +into solid carbon dioxide which has a temperature of -80° C. The evaporation of liquid ammonia by permitting it to escape into a pipe, under reduced pressure, is used on a large scale as a means of producing cold in cold storage @@ -9492,8 +9450,8 @@ evaporation.</p> <h4>Exercises</h4> -<p>1. How much heat is required (a) to melt 1 g. of ice at 0°C., (b) -to raise the temperature of the water resulting to 100°C., (c) +<p>1. How much heat is required (a) to melt 1 g. of ice at 0°C., (b) +to raise the temperature of the water resulting to 100°C., (c) to change this water to steam?</p> <p>2. If the water leaving a steam radiator is as hot as the steam @@ -9507,8 +9465,8 @@ weather?</p> <p>5. What are two advantages of the high heat of vaporization of water?</p> -<p>6. If the heat from 1 g. of steam at 100°C. in changing to water -and cooling to 0°C. could be used in melting ice at 0°C. how +<p>6. If the heat from 1 g. of steam at 100°C. in changing to water +and cooling to 0°C. could be used in melting ice at 0°C. how much ice would be melted?</p> @@ -9721,10 +9679,10 @@ equivalent of heat but also the heat produced by burning coal or gas; 1 lb. of average soft coal should produce about 12,600 B.t.u. Now since 778 ft.-lbs. are equivalent to one B.t.u. the energy produced when 2 lbs. of average soft -coal is burned is 778 × 12,600 × 2 = 19,605,600 ft.-lbs. +coal is burned is 778 × 12,600 × 2 = 19,605,600 ft.-lbs. In actual practice 2 lbs. of average soft coal burned will develop about 1 horse-power for 1 hour. 1 horse-power-hour -= 33,000 ft.-lbs. × 60 = 1,980,000 ft.-lbs. Now efficiency += 33,000 ft.-lbs. × 60 = 1,980,000 ft.-lbs. Now efficiency equals (work out)/(work in) 1,980,000/19,605,600 = 1/10 or 10 per cent.. This is the efficiency of a good steam engine. Ordinary ones require 3 lbs. of coal burned to each horse-power-hour @@ -9819,7 +9777,7 @@ in horse-power hours?</p> <p>7. How many B.T.U.'s can be obtained by burning 1/2 ton of bituminous coal?</p> -<p>8. when a pound of water is heated from 40°F. to 212°F., how many +<p>8. when a pound of water is heated from 40°F. to 212°F., how many foot-pounds of energy are absorbed by the water?</p> <p>9. How many loads of coal each weighing 2 tons, could be lifted @@ -9831,13 +9789,13 @@ foot-pounds of energy are absorbed by the water?</p> </div> <p>10. When 3 cu. ft. of water are used for a hot bath and the water -has been heated from 50°F. to 112°F., how many B.T.U.'s +has been heated from 50°F. to 112°F., how many B.T.U.'s have been absorbed by the water?</p> <p>11. If the average temperature of water at the surface of Lake -Michigan is 50°F., how many B.T.U.'s would be given off +Michigan is 50°F., how many B.T.U.'s would be given off by each cubic foot of water at the surface, if the temperature -of the water should drop 5°F.?</p> +of the water should drop 5°F.?</p> <p>12. In a cold storage plant carbon dioxide gas is used. The pipe<span class="pagenum"><a name="Page_222" id="Page_222">[Pg 222]</a></span> leading from the compression pump to the expansion valve passes @@ -9951,10 +9909,10 @@ carried. Suppose a gas engine produces 1 horse-power and uses 20 cu. ft. of gas an hour, what is its efficiency? 1 horse-power-hour -= 550 × 60 -× 60 = 1,980,000 ft.-lbs. += 550 × 60 +× 60 = 1,980,000 ft.-lbs. 20 cu. ft. of gas = 20 -× 600 × 778 = 9,336,000 +× 600 × 778 = 9,336,000 ft.-lbs.</p> <p>Efficiency = work out/work in = @@ -10013,12 +9971,12 @@ steamers are now driven by steam turbines.</p> amounts of heat can be secured from 1 cent's worth of each?</p> <p>2. What will it cost to heat 30 gallons of water (1 gal. of water -weighs about 8-1/3 lbs.) from 40°F. to 190°F. with coal costing +weighs about 8-1/3 lbs.) from 40°F. to 190°F. with coal costing $4.00 per ton and yielding 12,000 B.T U. per lb. if the heater has an efficiency of 50 per cent.</p> -<p>3. What will it cost to heat 30 gallons of water from 40°F. to -190°F. with gas at $0.80 per 1000 cu. ft. if the heating device +<p>3. What will it cost to heat 30 gallons of water from 40°F. to +190°F. with gas at $0.80 per 1000 cu. ft. if the heating device has an efficiency of 75 per cent.</p> <p>4. Construct a cardboard working-model showing the action of the @@ -10082,7 +10040,7 @@ the horse-power? (See problem 15.)</p> <p>Heat; sources (4), effects (5), units (2).</p> <p>Temperature; thermometer scales (3), absolute temperature, -9C°/5 + 32° = F°.</p> +9C°/5 + 32° = F°.</p> <p>Expansion; gases, Law of Charles (V<sub>1</sub>/V<sub>2</sub> = T<sub>1</sub>/T<sub>2</sub>), liquids, peculiarity of water, solids, coefficient of expansion, uses, results.</p> @@ -10567,8 +10525,8 @@ has since been shown to be correct. The north magnetic (or south-seeking) pole was found in 1831, by Sir James Ross in Boothia Felix, Canada. Its approximate present location as determined by Captain Amundsen in 1905 is<span class="pagenum"><a name="Page_239" id="Page_239">[Pg 239]</a><br /><a name="Page_240" id="Page_240">[Pg 240]</a></span> -latitude 70° 5´ N. and longitude 96° 46´ W. The south -magnetic pole is in latitude 72° S., longitude 155° 16´ E. +latitude 70° 5´ N. and longitude 96° 46´ W. The south +magnetic pole is in latitude 72° S., longitude 155° 16´ E. The north magnetic pole is continually changing its position. At present it is moving slowly westward.</p> @@ -10588,7 +10546,7 @@ variation is called <i>declination</i>. It is defined as the <i>angle between the direction of the needle and the geographical meridian</i>. Declination is due to the fact that the geographical and magnetic poles do not coincide. What is -meant by a declination of 90°? Lines drawn upon a map +meant by a declination of 90°? Lines drawn upon a map so as to pass through places of the same declination are called <i>isogonic</i> lines. The line passing through points where the needle points north, without declination, is @@ -10605,11 +10563,11 @@ equilibrium, that is, so as to remain balanced in any position in which it is left. Upon being magnetized and placed so that it can swing in a north and south plane, the north-seeking pole will now be found to be depressed, -the needle forming an angle of nearly 70° with the horizontal. +the needle forming an angle of nearly 70° with the horizontal. (See Fig. 185.) The position assumed by the needle indicates that the earth's magnetic field instead of<span class="pagenum"><a name="Page_241" id="Page_241">[Pg 241]</a></span> being horizontal in the United States <i>dips</i> down at an angle -of about 70°. Over the magnetic pole, the <i>dipping needle</i> +of about 70°. Over the magnetic pole, the <i>dipping needle</i> as it is called, is vertical. At the earth's equator it is nearly horizontal. <i>The angle between a horizontal plane and the earth's magnetic lines of force @@ -11280,7 +11238,7 @@ will now consider the conditions that produce the "flow" or "current" of electricity. Let two electroscopes stand near each other. Charge -one, <i>C´</i> (Fig. 206), strongly +one, <i>C´</i> (Fig. 206), strongly and charge the other slightly. If now a light stiff wire attached to a @@ -12011,11 +11969,11 @@ polarization or local action, (c) very low internal resistance, <div class="figleft" style="width: 200px;"> <img src="images/i_293.png" width="200" height="178" alt="" /> -<span class="caption"><span class="smcap">Fig. 222.</span>—The Leclanché cell, +<span class="caption"><span class="smcap">Fig. 222.</span>—The Leclanché cell, "wet" type.</span> </div> -<p><b>249. The Leclanché cell</b> is the one commonly used for +<p><b>249. The Leclanché cell</b> is the one commonly used for ringing door bells. It has two plates: one of zinc and the other of <i>carbon</i>. These are placed in a solution of sal ammoniac (Fig. 222). Take up the desirable qualities @@ -12039,7 +11997,7 @@ part of the time and is closed only occasionally; as in ringing door bells, operating telephones, and other devices whose circuits are usually open.</p> -<p><b>250. The Dry Cell.</b>—Many forms of Leclanché cells +<p><b>250. The Dry Cell.</b>—Many forms of Leclanché cells are made. One of these is called the <i>dry cell</i> (See Fig. 223.) In this cell the zinc plate is made into a jar or can and contains the other materials. At the center of the @@ -12052,13 +12010,13 @@ evaporation. The great advantage of this cell is that it may be used or carried in any position without danger of spilling its contents. Dry cells are often used to operate the spark coils of gas and -gasoline engines. The Leclanché +gasoline engines. The Leclanché cell described in Art. 249 is commonly known as the "wet cell."</p> <div class="figleft" style="width: 200px;"> <img src="images/i_294.png" width="200" height="254" alt="" /> -<span class="caption"><span class="smcap">Fig. 223.</span>—The Leclanché +<span class="caption"><span class="smcap">Fig. 223.</span>—The Leclanché cell, "dry" type.</span> </div> @@ -12078,7 +12036,7 @@ which are kept separated by a porous clay cup (Fig. 224). The zinc rod is kept in a solution of zinc sulphate contained in the porous cup. The copper plate is in a solution of copper sulphate filling the rest of the glass jar. Unlike -the Leclanché cell, this one must be kept upon a <i>closed +the Leclanché cell, this one must be kept upon a <i>closed circuit</i> to do its best work, as the two liquids mix when the circuit is open. Taking its @@ -12091,7 +12049,7 @@ plate. Therefore a uniform E.M.F. may be obtained from it, making it especially useful in laboratory experiments and tests. (c) Its resistance is considerable and (d) -it is more expensive to operate than the Leclanché. It is +it is more expensive to operate than the Leclanché. It is sometimes used upon closed circuits outside of laboratories<span class="pagenum"><a name="Page_277" id="Page_277">[Pg 277]</a></span> as in burglar and fire alarms, although in recent years, the storage battery is taking its place for these purposes.</p> @@ -12164,7 +12122,7 @@ cells.<span class="smcap">Fig. 228.</span>—A galvanoscope.</span> <h4>Important Topics</h4> -<p>1. Leclanché cells, (a), wet, (b), dry, construction, advantages, uses.</p> +<p>1. Leclanché cells, (a), wet, (b), dry, construction, advantages, uses.</p> <p>2. Daniell and gravity cells, construction, advantages, uses.</p> @@ -12759,7 +12717,7 @@ are given on p. 294.</p> <p><b>267. The ohm, the unit of resistance</b>, is defined by international agreement as follows: <i>An ohm is the resistance of a column of pure mercury, 106.3 cm. long with a cross-section -of a square millimeter and at a temperature of 0°C.</i></p> +of a square millimeter and at a temperature of 0°C.</i></p> <p>It should be noted that each of the four conditions affecting resistance is mentioned in the definition, viz., @@ -12834,7 +12792,7 @@ shown in Fig. 250.</p> <div class="center"> <table border="0" cellpadding="4" cellspacing="0" summary="Dimensions and Functions of Copper Wires"> -<tr><td rowspan="2">B. & S. gauge number</td><td colspan="2">Diameter</td><td rowspan="2">Circular mils</td><td rowspan="2">Sectional area in square millimeters</td><td rowspan="2">Weight and length, Density = 8.9, feet per pound</td><td rowspan="2">Resistance at 24°C., feet per ohm</td><td rowspan="2">Capacity in amperes</td></tr> +<tr><td rowspan="2">B. & S. gauge number</td><td colspan="2">Diameter</td><td rowspan="2">Circular mils</td><td rowspan="2">Sectional area in square millimeters</td><td rowspan="2">Weight and length, Density = 8.9, feet per pound</td><td rowspan="2">Resistance at 24°C., feet per ohm</td><td rowspan="2">Capacity in amperes</td></tr> <tr><td>Mils</td><td>Millimeters</td></tr> <tr><td align="right">0000</td><td align="right">460.000</td><td align="right">11.684</td><td align="right">211,600.00</td><td align="right">107.219</td><td align="right">1.56</td><td align="right">19,929.700</td><td align="right">312.0</td></tr> <tr><td align="right">000</td><td align="right">409.640</td><td align="right">10.405</td><td align="right">167,805.00</td><td align="right">85.028</td><td align="right">1.97</td><td align="right">15,804.900</td><td align="right">262.0</td></tr> @@ -13219,7 +13177,7 @@ given above, and we have <i>I</i> = 1.5/(0.5 + 2.5) = 1.5/3 = 0.5 ampere. Suppose again that these four cells were grouped in series with the same external resistance, substituting the values in the formula for cells in series we have <i>I</i> -= 4(1.5)/(4 × 2 + 2.5) = 6/10.5 = 0.57 ampere.</p> += 4(1.5)/(4 × 2 + 2.5) = 6/10.5 = 0.57 ampere.</p> <p><b>278. Volt-ammeter Method for Finding Resistance.</b>—Measurements of the resistance of conductors are often @@ -13698,32 +13656,32 @@ passing through the resistance between the points <i>M</i> and <i>N depends</i> upon three factors (1) the E.M.F. or <i>potential difference</i>; (2) the <i>current intensity</i> and (3) the <i>time</i>. The energy is measured by their product. That is, <i>electrical -energy</i> = <i>potential difference</i> × <i>current intensity</i> × <i>time</i>. +energy</i> = <i>potential difference</i> × <i>current intensity</i> × <i>time</i>. This represents the electrical energy in <i>joules</i>, or</p> <p> -Joules = volts × amperes × seconds, or<br /> -<i>j</i> = <i>E</i> × <i>I</i> × <i>t</i>.<br /> +Joules = volts × amperes × seconds, or<br /> +<i>j</i> = <i>E</i> × <i>I</i> × <i>t</i>.<br /> </p> <p>In the circuit represented in Fig. 268 the energy expended between the points <i>M</i> and <i>N</i> in 1 minute (60 seconds) is -8 × 2 × 60 = 960 joules.</p> +8 × 2 × 60 = 960 joules.</p> <p><b>291. Electric Power.</b>—Since power refers to the <i>time rate</i> at which work is done or energy expended, it may be computed by dividing the electrical energy by the time,<span class="pagenum"><a name="Page_317" id="Page_317">[Pg 317]</a></span> -or the <i>electrical power</i> = <i>volts</i> × <i>amperes</i>. The power +or the <i>electrical power</i> = <i>volts</i> × <i>amperes</i>. The power of 1 joule per second is called a <i>watt</i>. Therefore,</p> <p> -Watts = volts × amperes, or<br /> -Watts = <i>E</i> × <i>I</i>.<br /> +Watts = volts × amperes, or<br /> +Watts = <i>E</i> × <i>I</i>.<br /> </p> <p>Other units of power are the <i>kilowatt</i> = 1000 watts and the <i>horse-power</i> = 746 watts. In the example given -in Art. 290 the power of the current is 8 × 2 = 16 watts, +in Art. 290 the power of the current is 8 × 2 = 16 watts, or if the energy of the current expended between the joints <i>M</i> and <i>N</i> were converted into mechanical horse-power it would equal 16/746 of a horse-power. Electrical @@ -13834,26 +13792,26 @@ In other words, 1 joule will produce 1/4.2 or 0.24 calorie. Now the number of joules of electrical energy in an electric circuit is expressed by the following formula:</p> -<p>Joules = volts × amperes × seconds, or since 1 joule +<p>Joules = volts × amperes × seconds, or since 1 joule = 0.24 calorie,</p> <p> -Calories = volts × amperes × seconds × 0.24 or<br /> -<i>H</i> = <i>EI</i> × <i>t</i> × 0.24 (1)<br /> +Calories = volts × amperes × seconds × 0.24 or<br /> +<i>H</i> = <i>EI</i> × <i>t</i> × 0.24 (1)<br /> </p> -<p>By Ohm's law, <i>I</i> = <i>E</i>/<i>R</i> or <i>E</i> = <i>I</i> × <i>R</i>, substituting in equation +<p>By Ohm's law, <i>I</i> = <i>E</i>/<i>R</i> or <i>E</i> = <i>I</i> × <i>R</i>, substituting in equation (1) <i>IR</i> for its equal <i>E</i> we have</p> <p> -<i>H</i> = <i>I<sup>2</sup>R</i> × <i>t</i> × 0.24 (2)<br /> +<i>H</i> = <i>I<sup>2</sup>R</i> × <i>t</i> × 0.24 (2)<br /> </p> <p>Also since <i>I</i> = <i>E</i>/<i>R</i> substitute <i>E</i>/<i>R</i> for <i>I</i> in equation (1) and we have</p> <p> -<i>H</i> = <i>E<sup>2</sup></i>/<i>R</i> <i>t</i> × 0.24 (3)<br /> +<i>H</i> = <i>E<sup>2</sup></i>/<i>R</i> <i>t</i> × 0.24 (3)<br /> </p> <p><span class="pagenum"><a name="Page_320" id="Page_320">[Pg 320]</a></span></p> @@ -13861,7 +13819,7 @@ have</p> <p>To illustrate the use of these formulas by a problem suppose that a current of 10 amperes is flowing in a circuit having a resistance of 11 ohms, for 1 minute. The heat -produced will be by formula (2) = (10)<sup>2</sup> × 11 × 60 × 0.24 +produced will be by formula (2) = (10)<sup>2</sup> × 11 × 60 × 0.24 equals 15,840 calories.</p> <div class="figcenter" style="width: 400px;"> @@ -15266,11 +15224,11 @@ motion of wind. This mode of determining the speed of sound is not accurate. Other methods, more refined than the one just described have given accurate values for the speed of sound. The results of a number of experiments -show that, at the freezing temperature, 0°C., the +show that, at the freezing temperature, 0°C., the speed of sound in air is 332 meters or 1090 ft. a second. The speed of sound in air is affected by the temperature, increasing 2 ft. or 0.6 meter per second for each degree -that the temperature rises above 0°C. The speed decreases +that the temperature rises above 0°C. The speed decreases the same amount for each degree C. that the air is cooled below the freezing point. The speed of sound in various substances has been carefully determined. It is greater @@ -15312,29 +15270,29 @@ different musical instruments.</p> How do you explain this?</p> <p>4. How far away is a steamboat if the sound of its whistle is heard -10 seconds after the steam is seen, the temperature being 20°C.? +10 seconds after the steam is seen, the temperature being 20°C.? Compute in feet and in meters.</p> <p>5. How many miles away is lightning if the thunder is heard 12 -seconds after the flash in seen, the temperature being 25°C.?</p> +seconds after the flash in seen, the temperature being 25°C.?</p> <p>6. Four seconds after a flash of lightning is seen the thunder clap -is heard. The temperature is 90°F. How far away was the +is heard. The temperature is 90°F. How far away was the discharge?</p> <p>7. The report of a gun is heard 3 seconds after the puff of smoke -is seen. How far away is the gun if the temperature is 20°C.?</p> +is seen. How far away is the gun if the temperature is 20°C.?</p> <p>8. An explosion takes place 10 miles away. How long will it -take the sound to reach you, the temperature being 80°F?. -How long at 0°F.?</p> +take the sound to reach you, the temperature being 80°F?. +How long at 0°F.?</p> <p>9. How long after a whistle is sounded will it be heard if the distance -away is 1/4 mile, the temperature being 90°F.?</p> +away is 1/4 mile, the temperature being 90°F.?</p> <p>10. The report of an explosion of dynamite is heard 2 minutes after the puff of smoke is seen. How far away is the explosion -the temperature being 77°F.?</p> +the temperature being 77°F.?</p> <h3>(2) <span class="smcap">Waves<a name="FNanchor_N_14" id="FNanchor_N_14"></a><a href="#Footnote_N_14" class="fnanchor">[N]</a> and Wave Motion</span></h3> @@ -15496,7 +15454,7 @@ of its echo is the time that the sound takes to travel from its source to the reflecting body and back to the listener. Experiments have shown that the sensation of a sound persists about one-tenth of a second. Since the velocity -of sound at 20°C. is about 1130 ft. per second, during one-tenth +of sound at 20°C. is about 1130 ft. per second, during one-tenth of a second the sound wave will travel some 113 ft. If the reflecting surface is about 56 ft. distant a <i>short</i> sound will be followed immediately by its echo as it is heard one-tenth @@ -15533,10 +15491,10 @@ rarefaction.</p> <h4>Exercises</h4> <p>1. A hunter hears an echo in 8 seconds after firing his gun. How -far is the reflecting surface if the temperature is 20°C.?</p> +far is the reflecting surface if the temperature is 20°C.?</p> <p>2. How far is the reflecting surface of a building if the echo of -one's footsteps returns in 1 second at 10°C.?</p> +one's footsteps returns in 1 second at 10°C.?</p> <p>3. Why is it easier to speak or sing in a room than out of doors?</p> @@ -15550,7 +15508,7 @@ a person at the farther end of the building mentioned at the end of Art. 327?</p> <p>6. An echo is heard after 6 seconds. How far away is the reflecting -surface, the temperature being 70°F.?</p> +surface, the temperature being 70°F.?</p> <p>7. Why are outdoor band-stands generally made with the back curving over the band?</p> @@ -15563,7 +15521,7 @@ the moon? Explain.</p> <p>10. If a sunset gun was fired exactly at 6:00 <span class="smcap">P.M.</span> at a fort, at what time was the report heard by a man 25 miles away, if the temperature -was 10°C.?</p> +was 10°C.?</p> <h3>(3) <span class="smcap">Intensity and Pitch of Sounds</span></h3> @@ -15726,7 +15684,7 @@ the relation between the speed (<i>v</i>), wave length (<i>l</i>), and number of vibrations per sec. (<i>n</i>):</p> <p> -<i>v</i> = <i>l</i> × <i>n</i>, or <i>l</i> = <i>v/n</i><br /> +<i>v</i> = <i>l</i> × <i>n</i>, or <i>l</i> = <i>v/n</i><br /> </p> <p>that is, <i>the speed of a sound wave is equal to the number @@ -15759,14 +15717,14 @@ factors affecting intensity.</p> <i>do</i> has 120 vibrations.</p> <p>3. What is the wave length of the "A" of international concert -pitch at 25°C.? Compute in feet and centimeters.</p> +pitch at 25°C.? Compute in feet and centimeters.</p> <p>4. At what temperature will sound waves in air in unison with "Middle C" be exactly 4 ft. long?</p> <p>5. Explain the use of a megaphone.</p> -<p>6. What tone has waves 3 ft. long at 25°C.?</p> +<p>6. What tone has waves 3 ft. long at 25°C.?</p> <p>7. What is the purpose of the "sounding board" of a piano?</p> @@ -15776,7 +15734,7 @@ the two men?</p> <p>9. The speaking tone of the average man's voice has 160 vibrations per second. How long are the waves produced by him -at 20°C.?</p> +at 20°C.?</p> <h3>(4) <span class="smcap">Musical Scales and Resonance</span></h3> @@ -15803,7 +15761,7 @@ major scale.</p> <div class="center"> <table border="0" cellpadding="4" cellspacing="0" summary="Table of Musical Nomenclatures"> -<tr><td align="left">Name of note</td><td align="left"> C</td><td align="left">D</td><td align="left">E</td><td align="left">F</td><td align="left">G</td><td align="left">A</td><td align="left">B</td><td align="left">C´</td></tr> +<tr><td align="left">Name of note</td><td align="left"> C</td><td align="left">D</td><td align="left">E</td><td align="left">F</td><td align="left">G</td><td align="left">A</td><td align="left">B</td><td align="left">C´</td></tr> <tr><td align="left">Frequency in terms of "do"</td><td align="left"><i>n</i></td><td align="left">9/8<i>n</i></td><td align="left">5/4<i>n</i></td><td align="left">4/3<i>n</i></td><td align="left">3/2<i>n</i></td><td align="left">5/3<i>n</i></td><td align="left">15/8<i>n</i></td><td align="left">2<i>n</i></td></tr> <tr><td align="left">Intervals</td><td align="right">9/8</td><td align="right">10/9</td><td align="right">16/15</td><td align="right">9/8</td><td align="right">10/9</td><td align="right">9/8</td><td align="right">16/15</td></tr> <tr><td align="left">Name of note in vocal music</td><td align="left">do</td><td align="left">re</td><td align="left">mi</td><td align="left">fa</td><td align="left">sol</td><td align="left">la</td><td align="left">ti</td><td align="left">do</td></tr> @@ -15826,10 +15784,10 @@ vibration numbers</i> corresponding are 24, 30, 36. These in simplest terms have ratios of 4:5:6. Any three other tones with vibration ratios of 4:5:6 will also form a major triad. If the octave of the lower tone is added, the four make a -major chord. Thus: F, A, C´ (<i>fa</i>, <i>la</i>, <i>do</i>), 32:40:48, or 4:5:6, -also form a major triad as do G, B, D´ (<i>sol</i>, <i>ti</i>, <i>re</i>), 36:45:54, +major chord. Thus: F, A, C´ (<i>fa</i>, <i>la</i>, <i>do</i>), 32:40:48, or 4:5:6, +also form a major triad as do G, B, D´ (<i>sol</i>, <i>ti</i>, <i>re</i>), 36:45:54, or 4:5:6. Inspection will show that these three major -triads comprise all of the tones of the major scale D´ +triads comprise all of the tones of the major scale D´ being the octave of D. It is, therefore, said that the major scale is based, or built, upon these three major triads. The examples just given indicate the mathematical<span class="pagenum"><a name="Page_370" id="Page_370">[Pg 370]</a></span> @@ -15881,7 +15839,7 @@ C sharp and D flat become the same tone instead of two different tones. Such a scale is called a <i>tempered scale</i>. The tempered scale has 13 notes to the octave, with 12 equal intervals, the ratio between two successive notes -being the ¹²√2 or 1.059. That is, any vibration rate +being the ¹²√2 or 1.059. That is, any vibration rate on the tempered scale may be computed by multiplying the vibration rate of the preceding note by 1.059. While this is a necessary arrangement, there is some loss in perfect @@ -16009,9 +15967,9 @@ from your own experience out of school.</p> <p>5. An air column 2 ft. long closed at one end is resonant to what wave length? What number of vibrations will this sound have -per second at 25°C.?</p> +per second at 25°C.?</p> -<p>6. At 24°C. What length of air column closed at one end will be +<p>6. At 24°C. What length of air column closed at one end will be resonant to a sound having 27 vibrations a second?</p> <p>7. A given note has 300 vibrations a second. What will be the @@ -16519,7 +16477,7 @@ overtones? Give vibration numbers and pitch names or letters.</p> 400 ft. as at 100 ft.?</p> <p>5. A resonant air column 60 cm. long closed at one end will respond -to what rate of vibration at 10°C.?</p> +to what rate of vibration at 10°C.?</p> <p>6. Can you find out how the valves on a cornet operate to change the pitch of the tone?</p> @@ -16869,7 +16827,7 @@ surface 1 ft. away and at right angles to the light rays<span class="pagenum"><a is called a <b>foot-candle</b>. It is the unit of intensity of illumination. A 4-candle-power lamp, at a distance of 1 ft., produces 4 foot-candles. A 16-candle-power lamp -at a distance of 2 ft. also produces 4 foot-candles—(16 ÷ 2<sup>2</sup>).</p> +at a distance of 2 ft. also produces 4 foot-candles—(16 ÷ 2<sup>2</sup>).</p> <p>The intensity of illumination required for a good light for seeing varies with the conditions. Thus, for stage and @@ -16902,7 +16860,7 @@ may be shown, however, by an experiment.</p> <div class="figleft" style="width: 180px;"> <img src="images/i_414.png" width="180" height="236" alt="" /> -<span class="caption"><span class="smcap">Fig. 352.</span>—<i>B´</i> is as far +<span class="caption"><span class="smcap">Fig. 352.</span>—<i>B´</i> is as far back of the mirror as <i>B</i> is in front of it.</span> </div> @@ -17051,19 +17009,19 @@ way in which images are formed by a plane mirror may be illustrated by diagrams. Thus in Fig. 354, let <i>L</i> represent a -luminous body and <i>E</i> and <i>E´</i> +luminous body and <i>E</i> and <i>E´</i> two positions of the observer's eye. Take any line or ray as <i>LO</i> along which the light from -<i>L</i> strikes the mirror <i>O-O´</i>. It +<i>L</i> strikes the mirror <i>O-O´</i>. It will be reflected so that angle <i>LOP</i> equals angle <i>POE</i>. Similarly with any other ray, as -<i>LO´</i>, the reflected ray <i>O´E´</i> has -a direction such as that angle <i>L´O´E´</i> equals angle <i>P´O´E´</i>. +<i>LO´</i>, the reflected ray <i>O´E´</i> has +a direction such as that angle <i>L´O´E´</i> equals angle <i>P´O´E´</i>. Any other rays will be reflected in a similar manner, each of the reflected rays appearing to the eye to come -from a point <i>L´</i> behind the mirror.</p> +from a point <i>L´</i> behind the mirror.</p> <div class="figleft" style="width: 200px;"> <img src="images/i_419.png" width="200" height="261" alt="" /> @@ -17080,7 +17038,7 @@ sets up a series of waves spreading in all directions, so one may imagine a train of waves sent out by a luminous<span class="pagenum"><a name="Page_402" id="Page_402">[Pg 402]</a></span> body <i>L</i> (as in Fig. 355) to the mirror <i>MN</i>. These waves will be reflected from the mirror as if the source of light -were at <i>L´</i>. It is much simpler and more convenient to +were at <i>L´</i>. It is much simpler and more convenient to locate the position of the image of a point by the use of lines or "rays" (as in Fig. 354) than by the wave diagram (as in Fig. 355). In all <i>ray diagrams</i>, however, it should be @@ -17098,31 +17056,31 @@ rays.</p> <p><b>365. To locate the image of an object formed by a plane mirror</b> <i>requires</i> simply an application of the law of reflection. Thus in Fig. 356 let <i>AB</i> represent an object -and <i>MN</i> a plane mirror. Let <i>AA´</i> be a ray from <i>A</i> striking +and <i>MN</i> a plane mirror. Let <i>AA´</i> be a ray from <i>A</i> striking the mirror <i>perpendicularly</i>. It is therefore reflected back along the same line toward <i>A</i>. Let <i>AO</i> represent any other ray from <i>A</i>. It will be reflected along <i>OE</i> so that angle -<i>r</i> equals <i>i</i>. The intersection of <i>AC</i> and <i>OE</i> at <i>A´</i> behind +<i>r</i> equals <i>i</i>. The intersection of <i>AC</i> and <i>OE</i> at <i>A´</i> behind the mirror locates the image of the point <i>A</i>, as seen by reflection from the mirror. The triangles <i>ACO</i> and -<i>A´CO</i> may be proved equal by geometry. Therefore<span class="pagenum"><a name="Page_403" id="Page_403">[Pg 403]</a></span> -<i>A´C</i> equals <i>AC</i>. This indicates that <i>the image of a point +<i>A´CO</i> may be proved equal by geometry. Therefore<span class="pagenum"><a name="Page_403" id="Page_403">[Pg 403]</a></span> +<i>A´C</i> equals <i>AC</i>. This indicates that <i>the image of a point formed by a plane mirror is the same distance back of the mirror as the point itself is in front of it</i>. This principle -may be used in locating the image of point <i>B</i> at <i>B´</i>. Locating +may be used in locating the image of point <i>B</i> at <i>B´</i>. Locating the position of the <i>end points</i> of an image determines -the position of the whole image as <i>A´B´</i>.</p> +the position of the whole image as <i>A´B´</i>.</p> <div class="figcenter" style="width: 400px;"> <img src="images/i_421.png" width="400" height="259" alt="" /> -<span class="caption"><span class="smcap">Fig. 356.</span>—The image <i>A´B´</i> is as far back of the mirror <i>M N</i> as the +<span class="caption"><span class="smcap">Fig. 356.</span>—The image <i>A´B´</i> is as far back of the mirror <i>M N</i> as the object <i>A B</i> is in front of the mirror.</span> </div> <p><b>366. How the Image is Seen.</b>—Suppose the eye to be placed at <i>E</i>. It will receive light from <i>A</i> by reflection -as if it came from <i>A´</i>. Similarly light starting from <i>B</i> -reaches the eye from the direction of <i>B´</i>. There is nothing +as if it came from <i>A´</i>. Similarly light starting from <i>B</i> +reaches the eye from the direction of <i>B´</i>. There is nothing back of the mirror <i>in reality</i> that affects our sight, the light traveling only in the space in front of the mirror. Yet the action of the reflected light is such that it produces @@ -17845,8 +17803,8 @@ an axis at right angles to its equator to which are referred positions and distances, so a lens has a <i>principal axis</i> at right angles to its greatest diameter and along this axis are certain definite positions as shown in Fig. 383. Let -<i>MN</i> be the <i>principal axis</i> of a convex lens, <i>P</i> and <i>P´</i> are -<i>principal foci</i> on either side of the lens, <i>S</i> and <i>S´</i> are +<i>MN</i> be the <i>principal axis</i> of a convex lens, <i>P</i> and <i>P´</i> are +<i>principal foci</i> on either side of the lens, <i>S</i> and <i>S´</i> are <i>secondary foci</i>. These are at points on the principal axis that are twice as far from <i>O</i>, the center of the lens, as are the principal foci. In the formation of images by a convex @@ -17856,7 +17814,7 @@ lens, several distinct cases may be noticed:</p> its light is brought to a <i>focus</i> at <i>P</i>, or its <i>image is formed at P</i>. (B) As the <i>object approaches</i> the lens the <i>image gradually recedes</i> until the object and image are at <i>S</i> and -<i>S´</i>, <i>equally distant from O and of equal size</i> (as in Fig. 383). +<i>S´</i>, <i>equally distant from O and of equal size</i> (as in Fig. 383). The object and image are now said to be at the <i>secondary foci</i> of the lens. (C) As the <i>object moves from S to P</i> the image recedes, rapidly increasing in size until (D) when<span class="pagenum"><a name="Page_421" id="Page_421">[Pg 421]</a></span> @@ -18006,7 +17964,7 @@ the retina if the object is nearer than 10 in. (25 cm.).</p> <div class="figleft" style="width: 220px;"> <img src="images/i_442.png" width="220" height="138" alt="" /> <span class="caption"><span class="smcap">Fig. 387.</span>—The visual angle, <i>AOB</i> -is greater at <i>AB</i> than at <i>A´B´</i>.</span> +is greater at <i>AB</i> than at <i>A´B´</i>.</span> </div> <p><b>391. The Visual Angle.</b>—To examine objects carefully @@ -18032,7 +17990,7 @@ This is the principle of the magnifying glass used by watch-makers and of the <i>simple microscope</i>. The action of the latter is illustrated by Fig. 388. The convex lens forms a virtual, enlarged image -"<i>A´-B´</i>" of the object +"<i>A´-B´</i>" of the object "<i>A-B</i>" which it observed instead of the object itself.</p> @@ -18136,8 +18094,8 @@ the "objective" (<i>O</i>).</p> One called the <i>objective</i> is placed near the object to be viewed. This lens has a short focal length usually less than a centimeter. It forms a <i>real image</i> of the object. -<i>A´</i>-<i>B´</i>. The other lens, the <i>eyepiece</i> forms a virtual image -of this real image. <i>A´´</i>-<i>B´´</i>. (See Fig. 396.)</p> +<i>A´</i>-<i>B´</i>. The other lens, the <i>eyepiece</i> forms a virtual image +of this real image. <i>A´´</i>-<i>B´´</i>. (See Fig. 396.)</p> <p><span class="pagenum"><a name="Page_428" id="Page_428">[Pg 428]</a></span></p> @@ -18153,7 +18111,7 @@ light from distant stars the objective is made large, sometimes <div class="figcenter" style="width: 400px;"> <img src="images/i_446.png" width="400" height="190" alt="" /> <span class="caption"><span class="smcap">Fig. 396.</span>—Formation of an image by a microscope. <i>A</i>-<i>B</i> is the object. -<i>B´</i>-<i>A´</i> the real image formed by the "objective." <i>B´´</i>-<i>A´´</i> is the virtual +<i>B´</i>-<i>A´</i> the real image formed by the "objective." <i>B´´</i>-<i>A´´</i> is the virtual image formed by the eyepiece. The eye sees the virtual image.</span> </div> @@ -18917,17 +18875,17 @@ capacity of the condenser and the induction of the circuit.</p> <p>These oscillations pass through the primary of the oscillation transformer, inducing in the secondary, electric oscillations which -surge back and forth through the antennæ, or aerial wires, <i>A</i>. These +surge back and forth through the antennæ, or aerial wires, <i>A</i>. These oscillations set up the "wireless waves." The production of these waves is explained as follows: An electric current in a wire sets up a magnetic field spreading out about the conductor; when the current stops the field returns to the conductor and disappears. The -oscillations in the antennæ, however, have such a high frequency, +oscillations in the antennæ, however, have such a high frequency, of the order of a million a second, that when one surge of electricity sets up a magnetic field, the reverse surge immediately following sets up an opposite magnetic field before the first field can return to the wire. Under these conditions a succession of oppositely directed -magnetic fields are produced which move out from the antennæ<span class="pagenum"><a name="Page_451" id="Page_451">[Pg 451]</a></span> +magnetic fields are produced which move out from the antennæ<span class="pagenum"><a name="Page_451" id="Page_451">[Pg 451]</a></span> with the speed of light and induce electric oscillations in any conductors cut by them.</p></div> @@ -19040,11 +18998,11 @@ tube.</span> </div> <p><b>420. "X" Rays.</b>—In 1895, Professor -Röntgen of Wurtzburg, +Röntgen of Wurtzburg, Germany, discovered that when the cathode rays strike the walls of the tube or any solid within it they excite a form of invisible radiation. This radiation is called -Röntgen rays, or more commonly, "X" rays. Careful +Röntgen rays, or more commonly, "X" rays. Careful experiments show that they travel in straight lines, and that they can not be reflected or refracted as light waves are. They pass through glass and opaque objects such @@ -19361,12 +19319,12 @@ the tube; the variation corresponds to the vibrations of the transmitter diaphragm. This produces a surging current of the frequency of the sound waves in the primary of the transformer (<i>T</i>, Fig. 428). The secondary of this -transformer is connected to the antennæ (<i>A</i>) and the earth +transformer is connected to the antennæ (<i>A</i>) and the earth (<i>E</i>). By means of the transformer, rapid surgings are set -up in the antennæ and these surgings produce a continuous +up in the antennæ and these surgings produce a continuous stream of electromagnetic waves which goes out in space. (Like Fig. 426<i>C</i>.) These electromagnetic waves produce oscillations -in the antennæ of a receiving station. The antennæ +in the antennæ of a receiving station. The antennæ transmit the impulses to a <i>tube</i> (Fig. 427) which acts<span class="pagenum"><a name="Page_465" id="Page_465">[Pg 465]</a></span> as a <i>detector</i>, and makes possible the reproduction of the sound by an ordinary telephone receiver.</p> @@ -19599,7 +19557,7 @@ because it reduces the intensity of the current.</p> <div class="figcenter" style="width: 400px;"> <img src="images/i_490.png" width="400" height="251" alt="" /> <span class="caption"><span class="smcap">Fig. 434.</span>—Diagram showing graphically an alternating current with a -"lag" of 30° behind its electromotive force.</span> +"lag" of 30° behind its electromotive force.</span> </div> <p>Self-induction causes the current to <i>lag</i>, that is, the current @@ -19775,7 +19733,7 @@ Fig. 434), the product of volts and amperes gives only the <i>apparent power</i>, the ratio between true and apparent power depending on the amount of lag or lead. This ratio is called the power factor. In an a.-c. circuit, then, the -power equation is: watts = volts × amperes × power +power equation is: watts = volts × amperes × power factor, or power factor = true power/apparent power. The product of volts and amperes is the <i>apparent power</i> and is called volt-amperes in distinction from the true @@ -20555,7 +20513,7 @@ action.</p> <li>forms, <a href="#Page_416">416</a></li> </ul></li> -<li>Leclanché cell, <a href="#Page_275">275</a></li> +<li>Leclanché cell, <a href="#Page_275">275</a></li> <li>Lever, <a href="#Page_132">132</a></li> @@ -21147,7 +21105,7 @@ much greater mass.</p></div> <div class="footnote"> <p><a name="Footnote_G_7" id="Footnote_G_7"></a><a href="#FNanchor_G_7"><span class="label">[G]</span></a> The following formula is of assistance in computing <i>horse-power</i> in -problems: H. p. = (lbs. × ft.)/(550 × sec.).</p></div> +problems: H. p. = (lbs. × ft.)/(550 × sec.).</p></div> <div class="footnote"> @@ -21199,388 +21157,6 @@ electroplated.<br /> On page <a href="#Page_324">324</a>, Exercise number 8 was not used in the original. The exercises have not been renumbered.</p> - - - - - - - -<pre> - - - - - -End of the Project Gutenberg EBook of Physics, by -Willis Eugene Tower and Charles Henry Smith and Charles Mark Turton and Thomas Darlington Cope - -*** END OF THIS PROJECT GUTENBERG EBOOK PHYSICS *** - -***** This file should be named 40175-h.htm or 40175-h.zip ***** -This and all associated files of various formats will be found in: - http://www.gutenberg.org/4/0/1/7/40175/ - -Produced by Anna Hall, Albert László and the Online -Distributed Proofreading Team at http://www.pgdp.net (This -file was produced from images generously made available -by The Internet Archive) - - -Updated editions will replace the previous one--the old editions -will be renamed. - -Creating the works from public domain print editions means that no -one owns a United States copyright in these works, so the Foundation -(and you!) can copy and distribute it in the United States without -permission and without paying copyright royalties. Special rules, -set forth in the General Terms of Use part of this license, apply to -copying and distributing Project Gutenberg-tm electronic works to -protect the PROJECT GUTENBERG-tm concept and trademark. Project -Gutenberg is a registered trademark, and may not be used if you -charge for the eBooks, unless you receive specific permission. If you -do not charge anything for copies of this eBook, complying with the -rules is very easy. You may use this eBook for nearly any purpose -such as creation of derivative works, reports, performances and -research. They may be modified and printed and given away--you may do -practically ANYTHING with public domain eBooks. Redistribution is -subject to the trademark license, especially commercial -redistribution. - - - -*** START: FULL LICENSE *** - -THE FULL PROJECT GUTENBERG LICENSE -PLEASE READ THIS BEFORE YOU DISTRIBUTE OR USE THIS WORK - -To protect the Project Gutenberg-tm mission of promoting the free -distribution of electronic works, by using or distributing this work -(or any other work associated in any way with the phrase "Project -Gutenberg"), you agree to comply with all the terms of the Full Project -Gutenberg-tm License (available with this file or online at -http://gutenberg.org/license). - - -Section 1. General Terms of Use and Redistributing Project Gutenberg-tm -electronic works - -1.A. By reading or using any part of this Project Gutenberg-tm -electronic work, you indicate that you have read, understand, agree to -and accept all the terms of this license and intellectual property -(trademark/copyright) agreement. If you do not agree to abide by all -the terms of this agreement, you must cease using and return or destroy -all copies of Project Gutenberg-tm electronic works in your possession. -If you paid a fee for obtaining a copy of or access to a Project -Gutenberg-tm electronic work and you do not agree to be bound by the -terms of this agreement, you may obtain a refund from the person or -entity to whom you paid the fee as set forth in paragraph 1.E.8. - -1.B. "Project Gutenberg" is a registered trademark. It may only be -used on or associated in any way with an electronic work by people who -agree to be bound by the terms of this agreement. There are a few -things that you can do with most Project Gutenberg-tm electronic works -even without complying with the full terms of this agreement. See -paragraph 1.C below. There are a lot of things you can do with Project -Gutenberg-tm electronic works if you follow the terms of this agreement -and help preserve free future access to Project Gutenberg-tm electronic -works. See paragraph 1.E below. - -1.C. The Project Gutenberg Literary Archive Foundation ("the Foundation" -or PGLAF), owns a compilation copyright in the collection of Project -Gutenberg-tm electronic works. Nearly all the individual works in the -collection are in the public domain in the United States. If an -individual work is in the public domain in the United States and you are -located in the United States, we do not claim a right to prevent you from -copying, distributing, performing, displaying or creating derivative -works based on the work as long as all references to Project Gutenberg -are removed. Of course, we hope that you will support the Project -Gutenberg-tm mission of promoting free access to electronic works by -freely sharing Project Gutenberg-tm works in compliance with the terms of -this agreement for keeping the Project Gutenberg-tm name associated with -the work. You can easily comply with the terms of this agreement by -keeping this work in the same format with its attached full Project -Gutenberg-tm License when you share it without charge with others. - -1.D. The copyright laws of the place where you are located also govern -what you can do with this work. Copyright laws in most countries are in -a constant state of change. If you are outside the United States, check -the laws of your country in addition to the terms of this agreement -before downloading, copying, displaying, performing, distributing or -creating derivative works based on this work or any other Project -Gutenberg-tm work. The Foundation makes no representations concerning -the copyright status of any work in any country outside the United -States. - -1.E. Unless you have removed all references to Project Gutenberg: - -1.E.1. The following sentence, with active links to, or other immediate -access to, the full Project Gutenberg-tm License must appear prominently -whenever any copy of a Project Gutenberg-tm work (any work on which the -phrase "Project Gutenberg" appears, or with which the phrase "Project -Gutenberg" is associated) is accessed, displayed, performed, viewed, -copied or distributed: - -This eBook is for the use of anyone anywhere at no cost and with -almost no restrictions whatsoever. You may copy it, give it away or -re-use it under the terms of the Project Gutenberg License included -with this eBook or online at www.gutenberg.org/license - -1.E.2. If an individual Project Gutenberg-tm electronic work is derived -from the public domain (does not contain a notice indicating that it is -posted with permission of the copyright holder), the work can be copied -and distributed to anyone in the United States without paying any fees -or charges. If you are redistributing or providing access to a work -with the phrase "Project Gutenberg" associated with or appearing on the -work, you must comply either with the requirements of paragraphs 1.E.1 -through 1.E.7 or obtain permission for the use of the work and the -Project Gutenberg-tm trademark as set forth in paragraphs 1.E.8 or -1.E.9. - -1.E.3. If an individual Project Gutenberg-tm electronic work is posted -with the permission of the copyright holder, your use and distribution -must comply with both paragraphs 1.E.1 through 1.E.7 and any additional -terms imposed by the copyright holder. Additional terms will be linked -to the Project Gutenberg-tm License for all works posted with the -permission of the copyright holder found at the beginning of this work. - -1.E.4. Do not unlink or detach or remove the full Project Gutenberg-tm -License terms from this work, or any files containing a part of this -work or any other work associated with Project Gutenberg-tm. - -1.E.5. Do not copy, display, perform, distribute or redistribute this -electronic work, or any part of this electronic work, without -prominently displaying the sentence set forth in paragraph 1.E.1 with -active links or immediate access to the full terms of the Project -Gutenberg-tm License. - -1.E.6. You may convert to and distribute this work in any binary, -compressed, marked up, nonproprietary or proprietary form, including any -word processing or hypertext form. However, if you provide access to or -distribute copies of a Project Gutenberg-tm work in a format other than -"Plain Vanilla ASCII" or other format used in the official version -posted on the official Project Gutenberg-tm web site (www.gutenberg.org), -you must, at no additional cost, fee or expense to the user, provide a -copy, a means of exporting a copy, or a means of obtaining a copy upon -request, of the work in its original "Plain Vanilla ASCII" or other -form. Any alternate format must include the full Project Gutenberg-tm -License as specified in paragraph 1.E.1. - -1.E.7. Do not charge a fee for access to, viewing, displaying, -performing, copying or distributing any Project Gutenberg-tm works -unless you comply with paragraph 1.E.8 or 1.E.9. - -1.E.8. You may charge a reasonable fee for copies of or providing -access to or distributing Project Gutenberg-tm electronic works provided -that - -- You pay a royalty fee of 20% of the gross profits you derive from - the use of Project Gutenberg-tm works calculated using the method - you already use to calculate your applicable taxes. The fee is - owed to the owner of the Project Gutenberg-tm trademark, but he - has agreed to donate royalties under this paragraph to the - Project Gutenberg Literary Archive Foundation. Royalty payments - must be paid within 60 days following each date on which you - prepare (or are legally required to prepare) your periodic tax - returns. Royalty payments should be clearly marked as such and - sent to the Project Gutenberg Literary Archive Foundation at the - address specified in Section 4, "Information about donations to - the Project Gutenberg Literary Archive Foundation." - -- You provide a full refund of any money paid by a user who notifies - you in writing (or by e-mail) within 30 days of receipt that s/he - does not agree to the terms of the full Project Gutenberg-tm - License. You must require such a user to return or - destroy all copies of the works possessed in a physical medium - and discontinue all use of and all access to other copies of - Project Gutenberg-tm works. - -- You provide, in accordance with paragraph 1.F.3, a full refund of any - money paid for a work or a replacement copy, if a defect in the - electronic work is discovered and reported to you within 90 days - of receipt of the work. - -- You comply with all other terms of this agreement for free - distribution of Project Gutenberg-tm works. - -1.E.9. If you wish to charge a fee or distribute a Project Gutenberg-tm -electronic work or group of works on different terms than are set -forth in this agreement, you must obtain permission in writing from -both the Project Gutenberg Literary Archive Foundation and Michael -Hart, the owner of the Project Gutenberg-tm trademark. Contact the -Foundation as set forth in Section 3 below. - -1.F. - -1.F.1. Project Gutenberg volunteers and employees expend considerable -effort to identify, do copyright research on, transcribe and proofread -public domain works in creating the Project Gutenberg-tm -collection. Despite these efforts, Project Gutenberg-tm electronic -works, and the medium on which they may be stored, may contain -"Defects," such as, but not limited to, incomplete, inaccurate or -corrupt data, transcription errors, a copyright or other intellectual -property infringement, a defective or damaged disk or other medium, a -computer virus, or computer codes that damage or cannot be read by -your equipment. - -1.F.2. LIMITED WARRANTY, DISCLAIMER OF DAMAGES - Except for the "Right -of Replacement or Refund" described in paragraph 1.F.3, the Project -Gutenberg Literary Archive Foundation, the owner of the Project -Gutenberg-tm trademark, and any other party distributing a Project -Gutenberg-tm electronic work under this agreement, disclaim all -liability to you for damages, costs and expenses, including legal -fees. YOU AGREE THAT YOU HAVE NO REMEDIES FOR NEGLIGENCE, STRICT -LIABILITY, BREACH OF WARRANTY OR BREACH OF CONTRACT EXCEPT THOSE -PROVIDED IN PARAGRAPH 1.F.3. YOU AGREE THAT THE FOUNDATION, THE -TRADEMARK OWNER, AND ANY DISTRIBUTOR UNDER THIS AGREEMENT WILL NOT BE -LIABLE TO YOU FOR ACTUAL, DIRECT, INDIRECT, CONSEQUENTIAL, PUNITIVE OR -INCIDENTAL DAMAGES EVEN IF YOU GIVE NOTICE OF THE POSSIBILITY OF SUCH -DAMAGE. - -1.F.3. LIMITED RIGHT OF REPLACEMENT OR REFUND - If you discover a -defect in this electronic work within 90 days of receiving it, you can -receive a refund of the money (if any) you paid for it by sending a -written explanation to the person you received the work from. If you -received the work on a physical medium, you must return the medium with -your written explanation. The person or entity that provided you with -the defective work may elect to provide a replacement copy in lieu of a -refund. If you received the work electronically, the person or entity -providing it to you may choose to give you a second opportunity to -receive the work electronically in lieu of a refund. If the second copy -is also defective, you may demand a refund in writing without further -opportunities to fix the problem. - -1.F.4. Except for the limited right of replacement or refund set forth -in paragraph 1.F.3, this work is provided to you 'AS-IS' WITH NO OTHER -WARRANTIES OF ANY KIND, EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO -WARRANTIES OF MERCHANTIBILITY OR FITNESS FOR ANY PURPOSE. - -1.F.5. Some states do not allow disclaimers of certain implied -warranties or the exclusion or limitation of certain types of damages. -If any disclaimer or limitation set forth in this agreement violates the -law of the state applicable to this agreement, the agreement shall be -interpreted to make the maximum disclaimer or limitation permitted by -the applicable state law. The invalidity or unenforceability of any -provision of this agreement shall not void the remaining provisions. - -1.F.6. INDEMNITY - You agree to indemnify and hold the Foundation, the -trademark owner, any agent or employee of the Foundation, anyone -providing copies of Project Gutenberg-tm electronic works in accordance -with this agreement, and any volunteers associated with the production, -promotion and distribution of Project Gutenberg-tm electronic works, -harmless from all liability, costs and expenses, including legal fees, -that arise directly or indirectly from any of the following which you do -or cause to occur: (a) distribution of this or any Project Gutenberg-tm -work, (b) alteration, modification, or additions or deletions to any -Project Gutenberg-tm work, and (c) any Defect you cause. - - -Section 2. Information about the Mission of Project Gutenberg-tm - -Project Gutenberg-tm is synonymous with the free distribution of -electronic works in formats readable by the widest variety of computers -including obsolete, old, middle-aged and new computers. It exists -because of the efforts of hundreds of volunteers and donations from -people in all walks of life. - -Volunteers and financial support to provide volunteers with the -assistance they need, are critical to reaching Project Gutenberg-tm's -goals and ensuring that the Project Gutenberg-tm collection will -remain freely available for generations to come. In 2001, the Project -Gutenberg Literary Archive Foundation was created to provide a secure -and permanent future for Project Gutenberg-tm and future generations. -To learn more about the Project Gutenberg Literary Archive Foundation -and how your efforts and donations can help, see Sections 3 and 4 -and the Foundation web page at http://www.pglaf.org. - - -Section 3. Information about the Project Gutenberg Literary Archive -Foundation - -The Project Gutenberg Literary Archive Foundation is a non profit -501(c)(3) educational corporation organized under the laws of the -state of Mississippi and granted tax exempt status by the Internal -Revenue Service. The Foundation's EIN or federal tax identification -number is 64-6221541. Its 501(c)(3) letter is posted at -http://pglaf.org/fundraising. Contributions to the Project Gutenberg -Literary Archive Foundation are tax deductible to the full extent -permitted by U.S. federal laws and your state's laws. - -The Foundation's principal office is located at 4557 Melan Dr. S. -Fairbanks, AK, 99712., but its volunteers and employees are scattered -throughout numerous locations. Its business office is located at -809 North 1500 West, Salt Lake City, UT 84116, (801) 596-1887, email -business@pglaf.org. Email contact links and up to date contact -information can be found at the Foundation's web site and official -page at http://pglaf.org - -For additional contact information: - Dr. Gregory B. Newby - Chief Executive and Director - gbnewby@pglaf.org - - -Section 4. Information about Donations to the Project Gutenberg -Literary Archive Foundation - -Project Gutenberg-tm depends upon and cannot survive without wide -spread public support and donations to carry out its mission of -increasing the number of public domain and licensed works that can be -freely distributed in machine readable form accessible by the widest -array of equipment including outdated equipment. Many small donations -($1 to $5,000) are particularly important to maintaining tax exempt -status with the IRS. - -The Foundation is committed to complying with the laws regulating -charities and charitable donations in all 50 states of the United -States. Compliance requirements are not uniform and it takes a -considerable effort, much paperwork and many fees to meet and keep up -with these requirements. We do not solicit donations in locations -where we have not received written confirmation of compliance. To -SEND DONATIONS or determine the status of compliance for any -particular state visit http://pglaf.org - -While we cannot and do not solicit contributions from states where we -have not met the solicitation requirements, we know of no prohibition -against accepting unsolicited donations from donors in such states who -approach us with offers to donate. - -International donations are gratefully accepted, but we cannot make -any statements concerning tax treatment of donations received from -outside the United States. U.S. laws alone swamp our small staff. - -Please check the Project Gutenberg Web pages for current donation -methods and addresses. Donations are accepted in a number of other -ways including checks, online payments and credit card donations. -To donate, please visit: http://pglaf.org/donate - - -Section 5. General Information About Project Gutenberg-tm electronic -works. - -Professor Michael S. Hart is the originator of the Project Gutenberg-tm -concept of a library of electronic works that could be freely shared -with anyone. For thirty years, he produced and distributed Project -Gutenberg-tm eBooks with only a loose network of volunteer support. - - -Project Gutenberg-tm eBooks are often created from several printed -editions, all of which are confirmed as Public Domain in the U.S. -unless a copyright notice is included. Thus, we do not necessarily -keep eBooks in compliance with any particular paper edition. - - -Most people start at our Web site which has the main PG search facility: - - http://www.gutenberg.org - -This Web site includes information about Project Gutenberg-tm, -including how to make donations to the Project Gutenberg Literary -Archive Foundation, how to help produce our new eBooks, and how to -subscribe to our email newsletter to hear about new eBooks. - - -</pre> - +<div>*** END OF THE PROJECT GUTENBERG EBOOK 40175 ***</div> </body> </html> |