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alexgriva [62]
2 years ago
11

A person applies a force of 200 N over 1.5 m to a jack. The jack exerts a 1000-N force on a car a distance of 0.02 m. What is th

e efficiency rating of the jack? HINT: Use work formula first to solve for WORKout and again for WORKin...then use efficiency formula.
Chemistry
1 answer:
xz_007 [3.2K]2 years ago
6 0

Answer:

The efficiency rating of the jack is 0.067.

Explanation:

We have, a person applies a force of 200 N over 1.5 m to a jack. The jack exerts a 1000-N force on a car a distance of 0.02 m.

It is required to find the the efficiency rating of the jack. It is equal to the ratio of output work to the ratio of input work. So,

\eta=\dfrac{1000\times 0.02}{200\times 1.5}\\\\\eta=0.067

Thus, the efficiency rating of the jack is 0.067.

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SashulF [63]

Answer:

The vapor pressure in torr of the solution is 53.7

Explanation:

Vapor pressure lowering is the colligative property that is solved with this formula:

ΔP = P° . Xm

ΔP = Vapor pressure of pure solvent (P°) - vapor pressure of solution

Xm = mole fraction for solute.

Let's solve the mole fraction (moles of solute / total moles)

Total moles = moles of solute + moles of solvent

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Let's replace the data in the main formula.

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Vapor pressure of solution = 53.7 Torr

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