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natta225 [31]
3 years ago
15

Help me pleaseee, it’s due today: Two people push on a large gate as shown on the view from above in the diagram. If the moment

of inertia of the gate is 90 kgm2, what is the resulting angular acceleration of the gate?

Physics
1 answer:
Sladkaya [172]3 years ago
4 0

Answer:

the angular acceleration of the gate is approximately 1.61  \frac{rad}{s^2}

Explanation:

Recall the formula that connects the net torque with the moment of inertia of a rotating object about its axis of rotation, and the angular acceleration (similar to Newton's second law with net force, mass, and linear acceleration):

\sum \tau_1=I\,\alpha

In our case, both forces contribute to the same direction of torque, so we can add their torques up and get the net torque on the gate:

\tau_{net}=(20*2+30*3.5) \,N\,m=145\,\,N\,m

Now we use this value to obtain the angular acceleration by using the given moment of inertia of the rotating gate:

\sum \tau_1=I\,\alpha\\145\,\,N\.m=(90\,\,kg\,m^2)\,\alpha\\\alpha= \frac{145}{90} \frac{rad}{s^2} = 1.61\, \frac{rad}{s^2}

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2 years ago
A rigid, nonconducting tank with a volume of 4 m3 is divided into two unequal parts by a thin membrane. One side of the membrane
kondor19780726 [428]

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

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<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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