Physical ArithmeticMacmillan, 1885 - 357 sider |
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Resultat 1-5 av 84
Side 17
... feet of gas per hour is used on an average 6 hours per day during a year . Required the cost at 4s . 6d . per 1000 cubic feet . 5. If , for this country , we have on an average eighteen letters per head per annum , find the average ...
... feet of gas per hour is used on an average 6 hours per day during a year . Required the cost at 4s . 6d . per 1000 cubic feet . 5. If , for this country , we have on an average eighteen letters per head per annum , find the average ...
Side 68
... feet = link . 4. One metre contains 3.2809 English , and 3 · 1862 Prussian feet . By what fraction of an English inch does the Prussian exceed the English foot ? 5. Determine the height of Mont Blanc in old Parisian feet , its height ...
... feet = link . 4. One metre contains 3.2809 English , and 3 · 1862 Prussian feet . By what fraction of an English inch does the Prussian exceed the English foot ? 5. Determine the height of Mont Blanc in old Parisian feet , its height ...
Side 71
... feet , but its regulation equivalent is 6,080 feet . The common knot is the length of a minute on the mean meridian circumference ; it is estimated to be equivalent to 6,076 feet . 1 league 3 knots ; 1 degree = 60 knots . - The ...
... feet , but its regulation equivalent is 6,080 feet . The common knot is the length of a minute on the mean meridian circumference ; it is estimated to be equivalent to 6,076 feet . 1 league 3 knots ; 1 degree = 60 knots . - The ...
Side 77
... feet in the side , and the perpendicular height is 486 feet ; required the length of the slant side of one of the triangular faces . 14. A man 5 feet 4 inches in height standing at a distance of 52 feet from the base of an electric ...
... feet in the side , and the perpendicular height is 486 feet ; required the length of the slant side of one of the triangular faces . 14. A man 5 feet 4 inches in height standing at a distance of 52 feet from the base of an electric ...
Side 81
... feet . suming that its square was 8 Egyptian acres , the acre being the square of 100 royal cubits ; required the length of the royal cubit . 4862 foot28 Egyptian acres , 1 Egyptian acre = 1002 cubit2 , 4862 8 × 1002 foot2 = cubit2 ...
... feet . suming that its square was 8 Egyptian acres , the acre being the square of 100 royal cubits ; required the length of the royal cubit . 4862 foot28 Egyptian acres , 1 Egyptian acre = 1002 cubit2 , 4862 8 × 1002 foot2 = cubit2 ...
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acid acres amount ampere angle axis C.G.S. system C.G.S. unit Calculate Cambridge cent Centigrade centimetres College constant copper cross-section Crown 8vo cubic centimetres cubic feet cubic foot cubic inches deduce degrees denote density diameter distance dyne Edition electricity electromotive force ELEMENTARY equal equivalent EXAMPLES EXERCISE Fahr Fahrenheit Fcap feet long feet per sec feet per second foot-pounds force francs gallon grains grammes Hence horizontal horse-power hydrogen increment intensity of gravity interval kilogramme liquid litres magnetic mercury metre miles an hour miles per hour millimetre minute numerous Illustrations ohms oxygen pence plane pound poundals pressure Professor quantity radian radius reciprocal resistance SECTION shillings silver solid specific gravity specific heat speed square inch standard substance Suppose surface temperature thick TREATISE velocity vibrations volume water by deg weight wire yards
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