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Assoli18 [71]
3 years ago
5

A large truck breaks down out on the road and receives a push back to town by a small compact car.

Physics
1 answer:
Anton [14]3 years ago
5 0

Answer:

A True. It agrees with Newton's third law

3 True. The car pushes the truck and goes at constant speed

Explanation:

To examine the final answers, we must silver the solution of the problem, if we see Newton's third law, action and reaction, we see that the car pushes the truck the action, the truck must oppose this force with a force applied on the car of equal magnitude, but opposite direction

In view of the above, let's review the statements.

A True. It agrees with Newton's third law

B False. Violate Newton's third law

C False  violate Newton´s third law

D False. The force is exerted by the car not specifically by the engine

E Faults if no force is exerted the truck should stop due to friction

Second question

If the two vehicles move at the same speed, the resulting force on each of them must be zero

1 False. If the truck doesn't get nine, it can't go at cruising speed

2 False if the car is accelerating it cannot go at constant speed

3 True. The car pushes the truck and goes at constant speed

4 False. If the truck slows it should slow down

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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:

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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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