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Svetlanka [38]
4 years ago
10

A ball has 116.62 J of gravitational potential energy at a height of 85 m. What is the mass of the ball?

Physics
1 answer:
SashulF [63]4 years ago
8 0

Answer:0.14 kg

Explanation:

Gravitational potential energy=pe=116.62J

Height=h=85m

Acceleration due to gravity=g=9.8m/s^2

Mass=m

Pe=m x g x h

116.62=m x 9.8 x 85

116.62=m x 833

Divide both sides by 833

116.62 ➗ 833=(m x 833) ➗ 833

0.14=m

Mass=0.14 kg

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Explanation:

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             W = -nRT ln(\frac{V_{2}}{V_{1}})

Putting the given values into the above formula as follows.

        W = -nRT ln(\frac{V_{2}}{V_{1}})

             = - 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})

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or,         = 29.596 kJ       (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

         W = \frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}

              = \frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}

              = \frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}

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               P_{1}V_{1} = P_{2}V_{2}

or,       P_{2} = \frac{P_{1}V_{1}}{V_{2}}

                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})

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Hence, the required pressure is 11 atm.

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          P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}

or,       P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}

                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}

                    = 28.7 atm

Therefore, required pressure is 28.7 atm.

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