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Veseljchak [2.6K]
2 years ago
7

Hooke’s law states that the distance that a spring is stretched by hanging object varies directly as the mass of the object. If

the distance is 140 cm when the mass is 21 kg, what is the distance when the mass is 5 kg?
Physics
1 answer:
trasher [3.6K]2 years ago
8 0

Answer:

d_{2} = 33.33 cm

Explanation:

Given:

When mass, m_{1} =21 kg

          distance travelled is  d_{1}  = 140 cm

When mass, m_{2} =5 kg

         distance travelled is  d_{2}  = ?

Hooke's law state that within elastic limit, when an external force is applied to a body, the body gets deformed and when the force is released the gets back to its original form.

Therefore according to the question,

\frac{d_{1}}{m_{1}}=\frac{d_{2}}{m_{2}}

\frac{140}{21}=\frac{d_{2}}{5}

d_{2} = 33.33 cm

Distance travelled is 33.33 cm when mass is 5 kg.

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c) PE_{f} =0.5\ PE_{i}

Explanation:

A parallel plate capacitor with C_{o} that is connected to a voltage source V_{o} holds a charge of Q_{o} =C_{o} V_{o}. Then we disconnect the voltage source and keep the charge Q_{o} constant . If we insert a dielectric of \kappa=2 between the plates while we keep the charge constant, we found that the potential decreases as:

                                                     V=\frac{V_{o}}{\kappa}

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                                              C=\frac{Q}{V} =\kappa\frac{Q_{o}}{V_{o}}=\kappa\ C_{o}

The stored energy without the dielectric is

                                               PE_{i}=\frac{1}{2}\frac{Q_{o}^{2}}{C_{o}}=\frac{1}{2}C_{o}V_{o}^{2}

The stored energy after the dielectric is inserted is:

                                               PE_{f}=\frac{1}{2}\frac{Q^{2}}{C}=\frac{1}{2}CV^{2}

If we replace in the above equation the values of V and C we get that

                                         PE_{f}=\frac{1}{2}\kappa\ C_{o}(\frac{V_{o}}{\kappa})^{2}=\frac{1}{\kappa}(\frac{1}{2}C_{o}V_{o}^{2})

                                                   PE_{f} =\frac{PE_{i}}{\kappa}

Finally

                                                  PE_{f} =0.5\ PE_{i}

                                               

                                     

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