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KonstantinChe [14]
3 years ago
8

SIL. TUILE I JUMUWrite down your own examples for each of the different effects of forces.​

Physics
2 answers:
Harrizon [31]3 years ago
5 0

Answer:

magnetic force

Explanation:

when some one gets charged up by maybe static eletricity the person will sustain the energy and the energy will make a force field around the person which can and will attract paper and he will be attractive

zavuch27 [327]3 years ago
3 0

Explanation:

Examples for effect of force - example

1) Can change the state of an object(rest to motion/ motion to rest):For example, pushing a heavy stone in order to move it. 2) May change the speed of an object if it is already moving. ... 3) May change the direction of motion of an object

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An airplane engine starts from rest; and 2 seconds later, it is rotating with an angular speed of 300 rev/min. if the angular ac
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First of all, we need to convert the angular speed from rev/min into rev/s:

\omega_f=300 rev/min=5 rev/s

The angular acceleration is the variation of angular speed divided by the time:

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And this is constant, so we can use the following equation to calculate the angle through which the engine has rotated:

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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
Stells [14]

Answer:

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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3 years ago
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