What is the length a rubberband was stretched if it has a spring constant of 5700N/m and is currently holding 8600J OF POTENTIAL
energy?
1 answer:
Answer:
Explanation:
<u>Elastic Potential Energy </u>
Is the energy stored in an elastic material like a spring of constant k, in which case the energy is proportional to the square of the change of length Δx and the constant k.
Given a rubber band of a spring constant of k=5700 N/m that is holding potential energy of PE=8600 J, it's required to find the change of length under these conditions.
Solving for Δx:
Substituting:
Calculating:
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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,
- Here it is zero.
,
- As we know that,
- Thus, the final temperature of the system will be equal to the initial temperature,
- We found that the process is isothermal,
- Thus, the work done will be,
Where, R is the universal gas constant.
<h3>What is a reversible process?</h3>- Any process which can be made to proceed in the reverse direction is called reversible process.
- During which, the system passes through exactly the same states as in the direct process.
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.
Learn more about thermodynamic processes here:
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