204 CONNOLLY'S ARITHMETIC. The following I put in the Guernsey Times, a paper printed in Cambridge, Guernsey county, Ohio, July 2nd, 1825.] A REBUS, BY JAMES L. CONNOLLY. Two-fifths of a liquid that in Egypt is found, One-fourth of a fruit that grows not far from the ground, Ans. WASHINGTON. WINTER EVENING QUESTIONS. 1. Says John to Richard, I can place four 1's in such a manner, that they will make 3, 7, 8, 12, and even 20. 2. Three jealous husbands and their wives being ready to pass by night over a river, find at the waterside a boat which can carry but two persons at a time, and for want of a waterman, are obliged to row themselves over at several times. The question is, how these six persons shall pass by two and two, so that none of the three wives may be found in the company of one or two men, unless her husband be present? 6 3. A countryman in his journey, having a fox, a goose, and a peck of corn, came to a river, where it happened he could carry but one over at a time. Now as no two were to be left together, that would destroy each other, he was at his wits' end how to dispose of them; for,' says he, though the corn can't eat the goose, nor the goose the fox, yet the goose can eat the corn, and the fox the goose.' The question is, how must he carry them over, that they may not devour each other? 4. Suppose the nine digits to be placed in a quadrangular form, in what order must they stand, that any three figures in a right line may make just 15 P 5. Two merry companions are to have equal shares of eight gallons of whiskey, which are in a vessel containing exactly eight gallons; and to divide it equally, they have only two other vessels empty, one of which contains five gallons, and the other three. The question is, how will they divide the said eight gallons of whiskey between them, by the help of these three vessels only, so that they may have four gallons each? PART II. APPENDIX. OF PENDULUMS. To find what length pendulums will be, to swing any given time. RULE. Multiply the square of the seconds, in any given. time, by 39,2, and the product will be the length in inches. EXAMPLES. 1. Required the length of a pendulum that will swing quarter seconds? ,25,25×39,2-2,45 inc. Ans. 2. Required the length of a pendulum that will swing half seconds? ,5X,5×39,2 9,8 inc. Ans. 3. Required the length of a pendulum that will swing seconds ? 1x1x39,2=39,2 inc. Ans. 4. Required the length of a pendulum that will swing minutes ? 60×60×39,2=141120 inc. or 2 miles 1200 feet. Ans. 5. Required the length of a pendulum that will swing hours? 3600 sec. equal 1 hour; then 3600×3600×39,2 equal the inches in 8018 miles, and 1152 inches over. 6. What is the difference between the length of a pendulum that swings half seconds, and one that swings three seconds? 3×3×39,2—,5×,5×39,2=2872 feet. Ans. 7. How often will a pendulum of 94 inches vibrate in a second ? RULE. Divide the given length by 39,2, and the quotient will be the square of the time in seconds. 9-9,839,2,25 or 4 of a second. Ans.. OF THE LEVER. To find what weight may be raised or balanced by any given power. RULE. As the distance between the body to be raised or balanced, and the prop, is to the distance between the prop and the point where the power is applied, so is the power to the weight which it will raise or balance. 1. Suppose a man weighs 200 lbs. standing on the end of a lever 12 feet long, what weight will he balance on the other end, supposing the prop 1,5 feet from the weight? 12 feet equal the length of the lever. 1 feet equal the distance of the weight from the prop. 10 feet equal the distance from the prop to the man. As 1: 200 :: 101 2. Suppose a body weighing 50 lbs. rests on the end of a lever 8 feet long, what weight will that body raise or balance on the other end, supposing the prop 2,5 feet from the weight to be raised? 8-2,5=5,5 Then as 2,5 : 50 :: 5,5 to 110lbs. Ans. OF THE WHEEL AND AXLE-TREE. RULE. As the diameter of the axle-tree is to the diameter of the wheel, so is the power applied to the wheel to the weight suspended by the axle-tree. 1. I would have a windlass made in such a manner, that a pound applied to the wheel should be equal to 12lbs. suspended from the axle-tree. Now supposing the diameter of the axle-tree 6 inches, required the diameter of the wheel? Say inversely 12: 61 to 72 inc. diameter of the wheel. Ans. 2. Suppose the diameter of the wheel 72 inches, required the diameter of the axle-tree, so that a pound applied to the wheel may balance 12 lbs. on the axle-tree? Say inversely 1:72: 12 to 6 inc. diameter of axle-tree. Ans. S. Suppose the diameter of the axlè-tree to be 6 inches, and that of the wheel 72 inches, what power applied at the wheel will balance 12lbs. at the axle-tree? Say inversely as 6 :. 12 :: 72 to 1 lb. Ans. 4. Suppose the diameter of the wheel 72 inches, and that of the axle-tree 6 inches, what weight at the axle-tree will balance one pound at the wheel? Say inversely as 72 : 1 :: 6 to 12 lbs. Ans. To find the quantity of pressure against the sluice or bank which pens water. RULE for question 62, page 196. 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 623, the number of pounds Avoirdupois in a cubic foot of fresh water, and the product will be the number of pounds required. RULE for question 63, page 196. The perpendicular pressure of fluids on the bottom of vessels, is estimated by the area of the bottom, multiplied by the altitude of the fluid. RULE for question 64, page 196. Find the fall by the next question, which will produce the given velocity; multiply that height by 62,5 lbs. Avoirdupois, for clean river water, by 63 lbs. for dirty water, and by 64 lbs. for sea water. EXAMPLES. 1. The mean velocity of a stream is 12 feet per second, what is the perpendicular fall of the stream? 12×12=144÷64-24 feet, answer. 2. What is the velocity of water issuing from a head of 6 feet deep? 6×64=384, square root whereof is 19,5+ feet. Ans. 3. If the velocity of a stream, issuing through the bulkhead of a mill, be 18 feet per second, what head of water is there? 18×18=324÷÷64-5 feet. Ans. 4. A miller has a head of water 6 feet 3 inches above the sluice, how high must the water be raised above the opening, so that half as much again water may be discharged from the sluice in the same time? RULE. The quantity of water discharged from a hole in a vessel, is as the square root of the height of water above the opening. 6,252,5 and half as much again is 2,5+1,253,75 for the square root of the required depth: then 3,753,7514,0625 feet high. Ans. 5. I observed, while a stone was falling from a precipice, a string with a bullet at the end, which measured 25 inches to the middle of the ball, make five vibrations. What was the height from whence the stone fell? 25÷39,2,6377 and ✔,6377,7985 of a second, the time of one vibration, and ,7985X54 seconds nearly, the time of the stone's descent: then 4X416, and 16X16256 feet. Ans. RULE. Take a line of any length, and by the last ques tion find the time from the dropping of the stone, till you hear it strike the bottom. Multiply 73088, (equal to 16x 4x1142 feet, the distance sound moves in a second,) by the number of seconds, till you hear the stone strike the bottom," and to this product add 1304164, (equal to the square of 1142,) and from the square root of that sum take 1142, and divide the square of the remainder by 64, (that is, 16×4,) and the quotient will be the depth of the well in feet. Divide the depth by 1142, and the quotient will be the time of the sound's ascent, which being taken from the whole time, will leave the time of the stone's descent in seconds. EXAMPLES. 1. Suppose I drop a stone into a well, and a line with a plummet, which measured to the middle of the ball 25 inc. made five vibrations, before I heard the stone strike the bottom of the well, I demand the depth, time of the stone's descent, and of the sound's ascent? 2539,2=,6377 and ✓,6377,7985 of a second, and ,7985×5= 4 sec. nearly, to the hearing of the stone strike: then 73088×4+1304164— 1142=121,53 and 121,53 × 121,53 64 230,77 feet, the depth, and 230,77÷1142,2 of a second, the time of the sound's ascent, and 4—,2— 3,8 seconds, the time of the stone's descent. Ans. Depth 230,77 feet, time of the stone's descent 3,8 seconds, and the sound's ascent ,2 of a second. 2. A bullet dropped from the top of a building was found to reach the ground in 13 seconds: I demand the height of the building? RULE. The square root of the feet in the space fallen through, will ever be equal to 4 times the number of seconds the body has been falling: therefore multiply the time by 4, and the square of the product will be the space fallen through in the given time. 1 sec. =1,5×4=6, and 6×6=36 feet high. Ans. 3. How many feet will a body fall in 6 seconds? 6×4=24, and 24×24=576 feet. Ans. 4. If a ball, discharged from a gun, returns to the earth in 14 seconds, how high did it ascend? RULE. Ascending bodies are retarded in the same ratio that descending bodies are accelerated. The ball being half of its time in its ascent, therefore, 14+2=7, and 7x4-28, and 28x28-784 feet. Ans. |