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Unformatted text preview: 1.6 × 108 N/m 2
A
C) 4.80 × 1012 N
G) 8.70 × 105 N
F < (1.6 × 108 N/m 2 )(7.70 × 10−4 m 2 ) = 1.23 × 105 N
D) 3.00 × 103 N
H) 4.23 × 109 N
10. A force of 250 N is applied to a hydraulic jack piston that is 0.02 m in diameter. If
the piston that supports the load has a diameter of 0.15 m, approximately how much mass
can be lifted by the jack? Ignore any difference in height between the pistons.
A) 1400 kg
E) 5600 kg
2
B) 700 kg
F) 3300 kg
⎛ A2 ⎞
⎛ d2 ⎞
C) 250 kg
G) 990 kg
P = P2 , F1 / A1 = F2 / A2 , F2 = ⎜ ⎟ F1 = ⎜ ⎟ F1
1
⎝ A1 ⎠
⎝ d1 ⎠
D) 2800 kg
H) 1700 kg
2 2 F2 ⎛ d2 ⎞ F1 ⎛ 0.15 ⎞ 250 N
m=
=
=
= 1400 kg
g ⎜ d1 ⎟ g ⎜ 0.02 ⎟ 9.81 N/kg
⎝⎠
⎝
⎠
2 Form D Exam 3...
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This document was uploaded on 03/03/2014 for the course PHY 231 at Michigan State University.
 Summer '08
 smith
 Physics, Acceleration

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