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Maslowich
3 years ago
15

On a drive from one city to? another, victor averaged 39 mph. if he had been able to average 72 ?mph, he would have reached his

destination 11 hrs earlier. what is the driving distance between one city and the? other?
Physics
1 answer:
PilotLPTM [1.2K]3 years ago
5 0
Let the unknown distance be xmiles
x/39-x/72=11hr
72x-39x/2808=11hr
33x/2808=11
33x= 30888
x=936miles
U can substitue back to check
at speed of 72mph, he would need 936/72=13hrs
at speed of 39mph, he would need 936/39=24hr
the difference is 24-13=11
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A rocket powered sled accelerates a jet pilot in training straight forward from rest to 270 km/h in 12.1 seconds. Find:
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A parallel-plate capacitor stores charge Q. The capacitor is then disconnected from its voltage source, and the space between th
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The relationship between the initial stored energy PE_{i} and the stored energy after the dielectric is inserted PE_{f} is:

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}

The capacitance is modified as:

                                              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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Question 23 of 32
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