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3 years ago

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A 0.245 kg ball is thrown straight up from 2.07 m above the ground. Its initial vertical speed is 8.00 m/s. A short time later,

it hits the ground. Calculate the total work done by the force of gravity during that time.

1 answer:

iris [78.8K]3 years ago

3 0

Answer:

The work done by gravity is $4.975 \: Joules$

Explanation:

The data given in the question is :

Mass is $0.245 kg$

Height from ground is $2.07 m$

As we know , the work done is state function , it depends on initial and final position not on the path followed.

So, work done by gravity = change in potential energy

Work done = Initial potential energy - final potential energy

Insert values from question

Work done = $mass \times gravity \times (change \: in \: height)$

Work done = $0.245 kg \times 9.81 m/s^{2} \times 2.07 m$

So, work done = $4.975 Joules$

Hence the work done by gravity is $4.975 \: Joules$

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Argon gas enters steadily an adiabatic turbine at 900 kPa and 450C with a velocity of 80 m/s and leaves at 150 kPa with a veloc

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Answer:

Temperature at the exit = $267.3 C$

Explanation:

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To find the mass flow rate, we can apply the ideal gas laws to estimate the specific volume, from there we can get the mass flow rate.

Assuming Argon behaves as an Ideal gas, we have the specific volume $v_{1}$

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for Ideal gasses, the enthalpy change can be calculated using the formula

$h_{2}-h_{1}=C_{p}(T_{2}-T_{1})$

hence we have

$W_{out}= -m_{f}((C_{p}(T_{2}-T_{1}) + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})$

$250= -2.871((0.5203(T_{2}-450) + \frac{150^{2}}{2\times 1000} - \frac{80^{2}}{2\times 1000})$

<em>Note: to convert the Kinetic energy term to kilojoules, it was multiplied by 1000</em>

evaluating the above equation, we have $T_{2}=267.3C$

Hence, the temperature at the exit = $267.3 C$

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