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3 years ago

A 5.00 kg crate is on a 21.0° hill.

Physics

1 answer:

mojhsa [17]3 years ago

8 0

Answer: 4575N

Explanation:

For y component, W = mgcosø

W = 500×9.8cos21

W = 4574.54N

Find the diagram in the attached file

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Complete Question

The complete question is shown on the first uploaded image

Answer:

The workdone is $W = -177.275J$

Explanation:

From the question we are told that

The initial Volume is $Vi = 0.160 L$

The final volume is $V_f = 0.510L$

The external pressure is $P = 5.00 \ atm$

Generally the change in volume is

$\Delta V = V_f - V_i$

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Substituting values

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Now $1 \ L \cdot atm = 101.3J$

Therefore $W = -1.75* 101.3$

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