5. In how many seconds will a weight fall through a space of 10000 feet? RULE. Four times the number of seconds in which the body was in falling, will ever be equal to the square root of the space in feet, through which it has fallen. Therefore divide the square root of space fallen through by 4, and the quotient will be the space in which it has fallen. ✔10000=100, and 100÷4=25 sec. Ans. 6. How long must a bullet be in falling, to acquire a velocity of 192 feet per second? RULE. Four times the number of seconds in which a body has been falling, is equal to an eighth of the velocity in feet per second, acquired at the end of the fall. Therefore divide the given velocity by 8, and a fourth part of the quotient will be the answer. 192÷8=24, and 24÷4-6 seconds. Ans. 7. In what time will a musket-ball, dropped from the top of a steeple 625 feet high, come to the ground? 625-25, and 25÷4-6 seconds. Ans. 8. With what velocity will an iron-ball begin to descend per second, if raised 2000 miles above the surface of the earth? RULE. As the square of the earth's semidiameter is to 16 feet, so is the square of any other distance, counting from the earth's centre inversely, to the velocity with which it begins to descend per second. 4000 miles equal the earth's semidiameter, and 4000+2000 equal the distance from the earth's centre: then say inversely, As 4000X4000: 16 :: 6000×6000 to 7 feet per second. Ans. 9. How high must a ball be raised above the surface of the earth, to begin to descend with a velocity of 77 feet per second? As 16 4000x4000 :: 7 to 36000000 miles: then miles: then 6000-4000 (the earth's semidiameter) the surface of the earth, which is the answer. 36000000===6000 2000 miles above 10. If a musket-ball dropped from the top of a steeple 625 feet high in 6,25 seconds, required its mean velocity? RULE. Divide the space fallen through by the number of seconds it was in falling, and the quotient will be the mean velocity. 625-6,25-100 feet per second. Ans 11. If a ball falls through a space of 625 feet high in 64 seconds, with what velocity will it strike the ground? RULE. The velocity acquired at the end of any period, is equal to twice the mean velocity with which it passed during that period: or multiply the space fallen through by 64, the square root of which product is the velocity sought. 625-6,25-100; then 100×2-200 feet. Ans. 12. If a ball strikes the ground with a velocity of 50 feet per second, from what height did it fall? RULE. Divide the square of the velocity by 64, and the quotient will be the height required. 50×50-2500; then 2500÷64-89 feet. Ans. OF WATER. 13. There is a sluice, one end of which is 3 feet lower than the other: required the velocity of the stream per second? 3x64-192, and 192-13,85+ feet. Ans. The weight given to be raised by a balloon, to find its diameter. RULE. As the specific difference between common and inflammable air is to one cubic foot, so is any weight to be raised to the cubic feet contained in the balloon. Divide the cubic feet by ,5236, and the cube root of the quotient will be the diameter required to balance it with common air; but to raise it, the diameter must be something greater. EXAMPLE. I wish to construct a spherical balloon of sufficient capacity to ascend with three persons, weighing 5 cwt. 2 qrs. 24 lbs. and the balloon and a bag of sand to weigh 2 qrs. 4 lbs. Required the diameter of the balloon? By the table of specific gravities, a cubic foot of common air weighs 14 oz. Avoirdupois, and a cubic foot of inflammable air weighs three twenty-fifths of an ounce A voirdupois: then 14 equals 1,25, and three twenty-fifths equals 0,12: then 1,25 less 0,12 equals 1,13, the difference between a cubic foot of common and inflammable air. 5 cwt. 2 qrs. 24 lbs. equal 640 lbs. and 2 qrs. 4 lbs. equal 60 lbs. Then 640+60=700 lbs. and 700X1611200 ounces. Then as 1,13: 1 :: 11200 to 9941,5044 cubic feet: then 3/9911,5044÷ ,523626 feet diameter. Ans. The diameter of a balloon given to find what weight it is capable of raising. RULE. Multiply the cube of the diameter by ,5236, and the product will be the content in cubic feet. Then say, as one cubic foot is to the specific difference between common and inflammable air, so are the contents of the balloon to the weight it will raise. EXAMPLE. The diameter of a balloon is 2613 feet, what weight is it capable of raising? 261326,65: then 26,65×26,65×26,65X,5236=9911,5044 eubic feet: then 1,25-0,12=1,13, the difference. Then say, As 11,13 9911,5044 to 11199,882 oz. equal to 700 lbs. nearly. Ans. A Table of specific gravities, of several solid and fluid bodies: the second column contains their absolute weight, and the third their relative weight, in Avoirdupois ounces. OF THE SPECIFIC GRAVITIES OF BODIES. The specific gravities of bodies are as their densities or weight, bulk for bulk. A body is said to have two or three times the specific gravity of an other, when it contains two or three times as much matter in the same space. A body immersed in a fluid will sink, if it be heavier than its bulk of the fluid: if it be suspended therein, it will lose so much of what it weighed in the air, as its bulk of the fluid weighs. Hence all bodies of equal bulks, which will sink in fluids, lose equal weights when suspended therein; and unequal bodies lose in proportion to their bulks. How to find the specific gravities of bodies. If a body suspended under a scale by a hair, or any other way, at the end of the balance, be first counterpoised in air by a weight in the opposite scale, and then immersed in water, the equilibrium will be destroyed. If then so much weight be put into the scale to which the body is suspended, as will restore the equilibrium, without altering the weights in the opposite scale, that weight which restores the equilibrium will be equal to a quantity of water as big as the immersed body; and if the weight of the body in air be divided by what it loses in water, the quotient will shew how much that body is heavier than its bulk of water. Thus if a guinea, suspended in air, be counterbalanced by 129 grains in the opposite scale, and then, upon being immersed in water, it becomes so much lighter, as to require 7 grains to be put into the scale over it, to restore the equilibrium, it shews that a quantity of water, of equal bulk with the guinea, weighs 7,25 grains, by which dividing 129, (the weight of the guinea in air,) the quotient will be 17,793, which shews that the guinea is 17,793 times as heavy as its bulk of water. Thus may any piece of gold be tried, by weighing it first in air, and then in water; and if by dividing the weight in air by the loss in water, the quotient is 17,793, the gold is good; and if the quotient be 18, or between 18 and 19, the gold is very fine: but if it be less than 17, the gold is too much alloyed by being mixed with some other metal. If silver be tried in this manner, and found to be 11 times as heavy as water, it is very fine, and if it be 104 times as heavy, it is standard; but if it be of any less weight, compared with water, it is mixed with some lighter metal, such as tin, &c. It a piece of glass, brass, or lead, be immersed and suspended in different sorts of fluids, the different losses of weights therein will shew how much heavier it is than its bulk of the fluid, that fluid being lightest, in which the immersed body loses least of aerial weight. Common clear water, for common uses, is generally made a standard for comparing bodies, by which gravity may be represented by unity, or 1; or in case great accuracy be required, by 1,000, where three cyphers are annexed to give room to express the ratios of other gravities in larger numbers in the table. In doing this there is a twofold advantage, the first of which is, that by this means the specific gravities of bodies may be expressed to a greater degree of exactness; and the second, that the numbers of the table, considered as whole numbers, do also express the ounces Avoirdupois contained in a cubic foot of every sort of matter therein specified; because a cubic foot of common water is found to weigh very near 1000 ounces Avoirdupois, or 623 lbs. How to try spirituous liquors. A cubic inch of good brandy, rum, or other proof spirits, weighs 234 grains; therefore if a true cubic inch of any metal weighs 234 grs. less in spirits than in air, it shews that the spirits are proof; and if it loses less of its aerial weight in spirits, they are above proof; but if it lose more, they are under proof; for the better the spirits are, the lighter they are, and the worse the heavier. Or let any fluid of sufficient specific gravity be weighed first in air, next in water, and then in any other liquid. From its weight in air take its weight in water, and the remainder is the weight of its bulk of water. From its weight in air take its weight in the other liquid, and the remainder is the weight of the same quantity of that liquid. Divide the weight of this quantity of liquid by the weight of the same quantity of water, and the quotient will be the specific gravity of the liquid. All bodies expand with heat, and contract with cold, some more, and some less than others; therefore the specific gravities of bodies are not precisely the same in summer as in winter. EXAMPLES. 1. What proportion of rectified spirits of wine must be mixed with water, to make proof spirits, the specific gravity of rectified spirits being 850, that of proof spirits 925, and that of water 1000? 925 { 1000 850 75 75 An equal quantity of each. 2. What proportional weight of rectified spirits of wine and water must be mixed, to make proof spirits, the specific gravities as before ? 1000-850=1; or 100029, that is, as 20 to 17. 850 3. What is the specific gravity of best French brandy, consisting of six parts (measure) of rectified spirits of wine, and three parts water? 850×6=5100 1000×3=3000 Then 5100+3000-8100, and 6+3=9, and 8100-9-900, the specific gravity. 4. A storekeeper has 50 gallons of rum, whose specific gravity is 900: how much water must he add to reduce it to standard proof? 925 {1000) 25 water 75 rum As 75: 25: 50 to 16 gallons of water to be added. Ans. To find the magnitude of any body, when the weight is known. RULE. Divide the weight by the specific gravity, and the quotient will be the magnitude sought. EXAMPLE. What is the magnitude of several fragments of brick, whose weight is 14 ounces? 14+2000=,007 of a cubic foot, and ,007x1728=12,096 cubic inches. Ans, To find the specific gravity of any body, when the magnitude is known. RULE. Divide the weight by the magnitude, and the quotient will be the specific gravity. EXAMPLE. Suppose a piece of ash to contain 8 cubic feet, and weighs 6400 ounces, what is the specific gravity? 6400÷8=800, the specific gravity. Ans. |