1.25-12=1·13oz. difference. And 160x4+60=700=11200. oz. cub. foot. oz. cub. feet. As 1.13: 1 :: 11200: 9911-5044. And 9911.5044 3 •5236 =26.65 feet, [Ans. Given the diameter of a balloon, to find what weight it is capable of raising. 1. Multiply the cube of the diameter by 5236, and the product will be the content in cubic feet. 2. As one cubick foot is to the specifick difference between common and inflammable air; so is the content of the balloon to the weight it will raise. 114. The diameter of a balloon is 26-65 feet: What weight is it capable of raising? 26.65×26-65×26-65X-5236=9911·4+cubick feet. And cub. feet. oz. cub. foot.. oz. nearly. If the magnitude of any body be multiplied by its specifick gravity, the product will be its absolute weight. 115. What weight of lead will cover a house, the area of whose roof is 6000 feet, and the thickness of the lead of a foot? 6000×50 cub, feet, and its specifick gravity 11325×50= tons. cwt. qrs. lbs. oz. 566250 ounces=15 15 3 26 10 Ans. To find the magnitude of any thing when the weight is known. Divide the weight by the specifick gravity in the Table, and the quotient will be the magnitude sought. 116. What is the magnitude of several fragments of clear glass, whose weight is 13 ounces? 13-2600-005 of a cubick foot, and •005×1728=8.640 cubick inches, Ans. Having the magnitude and weight of any body given, to find its spe cifick gravity. Divide the weight by the magnitude, and the quotient will be the specifick gravity. 117. Suppose a piece of marble contains 3 cubick feet, and weighs 13531 or 21656 ounces: What is the specifick gravity? 21656-82707 the specifick gravity required, as by the Table. To find the quantity of pressure against the sluice or bank, which pens water. Multiply the area of the sluice, under water, by the depth of the centre of gravity, (which is equal to half the depth of the water) in feet, and that product again by C2 (the number of pounds Avoir dupois in a cubick foot of fresh water) or by 64-4 (the Avoirdepois weight of a cubick foot of salt water) and the product will be the number of pounds required. 118. Suppose the length of a sluice or floom be 30 feet, the width at bottom 4 feet, and the depth of the water 4 feet; what is the pressure against the side of the sluice? 30×4=120 feet the area of the bottom, and 120×2 (the depth of the centre of gravity) gives 240 cubick feet, and 240×62-5= 15000=6T. 13cwt. 3qrs. 20 Ans. The perpendicular pressure of fluids on the bottoms of vessels is estimated by the area of the bottom multiplied by the altitude of the fluid. 119. Suppose a vessel 3 feet wide, 5 feet long, and 4 feet high, what is the pressure on the bottom, it being filled with water to the brim? 3×5=15 square feet, the area of the bottom, and 15×4=60 cubick feet, and 60×62·5=3750=33 cwt. 1 qr. 26. THE USE OF THE BAROMETER. The Barometer is so formed, that a column of quicksilver is sup ported within it to such a height as to counterbalance the weight of a column of air, of an equal diameter, extending from the barometer to the top of the atmosphere. 120. At the surface of the earth, the height of this column of quicksilver is, at an average, almost 30 inches; when the barometer is at that height; what is the pressure of atmosphere on a square foot, and on the surface of a man's body, estimated at 14 square feet? As the cubick foot of quicksilver is 13600 ounces, Avoirdupois, and as the height in the barometer, is 2.5 feet, therefore 13600×2.5 =34000 ounces, 2125 pounds on a square foot; and 2125×14= 29750 pounds on a man's body. 121. If the mercury in a barometer, at the bottom of a tower, be observed to stand at 30 inches, and, on being carried to the top of it, be observed at 29-9 inches: What is the height of the tower? Divide 13600, the specifick gravity of quicksilver, by 1.25, the specifick gravity of air, and the quotient will be the height of the tower, in tenths of an inch. 13600 1.25 10880 10880 tenths, and ———=1088 inch.=903 feet Ans. 10 The number of feet, in height, of the atmosphere, corresponding with of an inch on the barometer is variable, depending on the temperature and density of the atmosphere. The variation, depending on the temperature, is shewn in the fol lowing Table, calculated for every 5 degrees, from 32 to 80, Fahrenheit's Thermometer, from whence it may be easily calculated, for the intermediate degress by allowing of a foot for each degree. 100 TABLE. › Thermo. Feet. The altitude, thus found; will be to the altitude cor32°186-86 rected for the density of the air, inversely, as the 35 87-49 mean height of the barometer, at the two stations, is 40 88.54 to 30 inches; therefore, 45 89-60 RULE.Multiply the mean height corresponding 5090-66 to the mean temperature of the two barometers 55 91-72 (found in the Table) by the tenths of an inch in the 60 92.77 difference of the two barometers, and this product by 65 93.82 30; divide this last product by the mean height of the two barometers, and the quotient will be the answer, or height required, with the errour of a few feet only, if the height be less than a mile.* 70 94-88 75 95 93 80 96-99 122. At the first station, suppose the barometer to stand at 29, and the thermometer at 60; at the second station, the barometer at 28, and the thermometer at 40: What is the height of the 2d station or the distance between the two places of observation? Barometer. sum=28.5 mean height of the two barometers. 29 28 Difference 1=10 tenths of an inch. Thermometer. First station =60 Second station =40 1)100 50 mean height of the two thermometers, against which, in the Table you will find 90-66, the mean temperature of the two barometers. Now, according to the rule 90 66×10×30÷ 28.5 954-3 feet, the Answer, nearly. * Leth = mean height of the barometer at its two stations, (or of two barometers, one at each station) in inches; d= difference of the two barometers in tenths of an inch; and n = number from the Table answering to the mcan tem30dn perature of the two thermometers accompanying the barometer, then altitude required nearly. the h The Act of Congress of April 29, 1816, regulating the currency within the United States of the gold coins of Great Britain, France, &c. enacted, That, of the gold coins of Great Britain and Portugal, 27 grs. 100 cents, or 1 pwt.=888 cents; do. 37 do.. do. do. =84 Of France, 27 grs.do. Spain, 28grs. do. Crowns of France, weighing 449grs. 110cents, or 1oz.=1178. Five franc pieces, weighing 386grs.=93.3 cts. or 1oz. 116 cts. The Spanish dollar, weighing not less than 415grs.=100 cents. The Federal dollar is to be of the weight and purity of the Spanish dollar; but, it is to weigh 416grs. and to contain 3711grs. of pure silver. The pound Sterling of England is $444 in the United States. The dollar is reckoned, therefore, at four shillings and six pence sterling; but in 1820, four shillings and six pence of English silver coin contained only 3631grs. of pure silver, being 8grs. less than is contained in the Federal dollar. One pound Troy weight of Standard Gold in England, contains 5280grs. of pure gold, and is coined into £46 14s. 6d. or 11214 pence. Hence a pound sterling contains 113-0016grs. of pure gold. The Eagle contains 247-5grs. of pure gold, and hence the pound sterling is worth in gold $4 56-572, and hence the dollar is worth in English gold 4s. 4-5656d. One pound Troy of Standard silver in England contains 5328grs. of pure silver, and is coined into 66 shillings or 792 pence. As the dollar contains 3711grs. of pure silver, the dollar is worth in English silver 4s. 7·1858d. In a pound sterling there is 1614 545grs. of pure silver, which is equal in silver to $4·34 3943. Taking the mean of the values of the dollar and the pound sterling in gold and silver, The value of the dollar is 4s 5.8757d. sterling. And the value of the pound sterling is $4.45-7331. This mean value of the dollar and the pound sterling is very near the values at which they are commonly estimated. [See "Report of the Secretary of State upon Weights and Measures," appendix C. to Congress, Feb. 1821.] The standard price of gold in England, is £3 17s. 101d. an oz. and of silver 5s. 2d. an oz. The standard weight of the English guinea is 5pwt. 9grs.; but it usually weighs 5pwt. 8grs. The standard coin of France is to contain one tenth of alloy, and the standard value of gold to silver is 15 to 1. |