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skelet666 [1.2K]
3 years ago
14

The equation used to predict the theoretical period Ty of a simple pendulum assumes a small amplitude of oscillation. A student

builds a pendulum by attaching one end of a string of known length to the ceiling and the other end to a small solid sphere. The sphere is pulled back so the string makes a 5 angle with the vertical and released from rest with a theoretical amplitude of Oscillation At The student measures the resulting experimental period Tg and after the sphere makes several oscillations, measures the experiment oscillation. In the absence of air resistance, T T and A. Which of the following best explains how air resistance affects the results of the experiment?
A) with resistance because the sphere will speed up
B) With air resistance As , because although the sphere slows down, it also will travel for a greater time
C) With resistance because the sphere will travel a greater distance
D) With air resistance, 7>7y, because the sphere will slow down
E) with resistance because although the sphere slows down, also will travel shorter distance
Physics
1 answer:
astra-53 [7]3 years ago
4 0

Answer:

The answer is "Choice E".

Explanation:

In this situation the option e is right because its resistance decreases through time, however, the time is the same for the same reason, whereas the sphere deteriorates, somehow it travels shorter distances however if the air resistance becomes are using the amplitude of movement declines, that's why other choices were wrong.

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Answer:

The current across the resistance is 0.011 A.

Explanation:

Total resistance, R = 25 ohms

Total current, I = 100 mA = 0.1 A

Let the voltage is V.

By the Ohm's law

V = I R

V = 0.1 x 25 = 2.5 V

Now the resistance is R' = 220 ohm

As they are in parallel so the voltage is same. Let the current is I'.

V = I' x R'

2.5 = I' x 220

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No.

Since repeated measurements are taken and the average and 95% confidence interval are calculated, the possibility of the lack of agreement being a random error has been minimized or even eliminated.

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Random errors can result from the instrument not being precise or from mistakes by the researcher.

Random errors can be minimized by taking multiple readings and averaging the results.

Since repeated measurements are taken and the average and 95% confidence interval are calculated, the possibility of the lack of agreement being a ransom error has been minimized.

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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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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}

                                               

                                     

5 0
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