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Flauer [41]
3 years ago
10

A projectile is fired horizontally from a gun that is 45.0m above flat ground, emerging from the gun with a speed of 250m/s? How

long does the projectile remain in the air? At what horizontal distance from the firing point does it strike the ground? What is the magnitude of the vertical component of its velocity as it strikes the ground?

Physics
1 answer:
Usimov [2.4K]3 years ago
5 0

Answer:

The magnitude of the vertical component of its velocity as it hits the ground is 30 m/s. The horizontal distance from the firing point is 750 meters.

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The final temperature of the system will be equal to the initial temperature, and which is 373K. The work done by the system is 409.8R Joules.

To find the answer, we need to know about the thermodynamic processes.

<h3>How to find the final temperature of the gas?</h3>
  • Any processes which produce change in the thermodynamic coordinates of a system is called thermodynamic processes.
  • In the question, it is given that, the tank is rigid and non-conducting, thus, dQ=0.
  • The membrane is raptured without applying any external force, thus, dW=0.
  • We have the first law of thermodynamic expression as,

                                dU=dQ-dW

  • Here it is zero.

                                  dU=0,

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                             dU=C_pdT=0\\\\thus,  dT=0\\\\or , T=constant\\\\i.e, T_1=T_2

  • Thus, the final temperature of the system will be equal to the initial temperature,

                          T_1=T_2=100^0C=373K

<h3>How much work is done?</h3>
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                               W=RT*ln(\frac{V_2}{V_1} )=373R*ln(\frac{4}{\frac{4}{3} })\\ \\W=409.8R J

Where, R is the universal gas constant.

<h3>What is a reversible process?</h3>
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Thus, we can conclude that, the final temperature of the system will be equal to the initial temperature, and which is 373K. The work done by the system is 409.8R Joules.

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Answer:

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