Solutions of Problems in Gage's Elements of Physics: Aslo a General Review ...

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Solutions of Problems in Gage's Elements of Physics: Aslo a General Review ...
Alfred Payson Gage
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Q. 3. The pressure on the top is nothing ; on the bottom it is 25 x 20x 15«= 7500«; on each of the sides it is 25 X 15 X 7.5« = 2812.5«; and on the ends, 20 X 15 x 7.5« = 2250«.
Q. 4. The additional pressure will be 100« for every 4'**" of area on the inner surface. The area of the bottom is 500**"" ; the additional pressure is, therefore, — Xl00« = 12,500«.
4 This, also, is evidently the pressure on the top. The additional pressure on each of the sides is 375 ^ XlOO« = 9375«; 4 and on each of
...the ends, 300 XlOO« = 75008.
4 182 soiiUTroKS to problems.
Page 69* Q. 5, The total prGsaure od the bottoQi ia 7500* + 12,500*= 20,000^; ttiat on tUo top, 12,500b ; on each side, 12,187.5«; an eaoli eud, Lh7r>0^ Q, 6, The anj9wer& to Q, 3 wonkl be 13.6 times greater: viz., 102,000* ou the bottom ; 38,250* on each Bide, and 30,600* nflidering Q, i, it makes no difference wliether mercniy or water h used. In the case of mercnry^ llic total pressure on the bottcmi is evidently 102,000*+ 12.500*= lU, 500'* ; on the top, 12,500*; ou each side, 38,250* + 9375* — 47,C25*; and on eaeh end, 30.


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