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erastovalidia [21]
3 years ago
10

As in the video, we apply a charge +Q to the half-shell that carries the electroscope. This time, we also apply a charge –Q to t

he other half-shell. When we bring the two halves together, we observe that the electroscope discharges, just as in the video. What does the electroscope needle do when you separate the two half-shells again?As in the video, we apply a charge + to the half-shell that carries the electroscope. This time, we also apply a charge – to the other half-shell. When we bring the two halves together, we observe that the electroscope discharges, just as in the video. What does the electroscope needle do when you separate the two half-shells again?It deflects more than it did at the end of the video.It does not deflect at all.It deflects the same amount as at end of the video.It deflects less than it did at the end of the video.
Physics
2 answers:
frez [133]3 years ago
8 0

Answer:It does not deflect at all.

Explanation:

Andreyy893 years ago
4 0

Answer:

Explanation:

When the positively charged half shell is brought in contact with the electroscope, its needle deflects due to charge present on the shell.

When the negatively charged half shell is brought in contact with the positively charged shell , the positive and negative charge present on each shell neutralises each other  .So both the shells lose their charges .The positive half shell also loses all its charges

When we separate the half shells , there will be no deflection  in the electroscope because both the shell have already lost their charges and they have become neutral bodies . So they will not be able to produce any deflection in the electroscope.

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A bat at rest sends out ultrasonic sound waves at 46.2 kHz and receives them returned from an object moving directly away from i
murzikaleks [220]

Answer:

f" = 40779.61 Hz

Explanation:

From the question, we see that the bat is the source of the sound wave and is initially at rest and the object is in motion as the observer, thus;

from the Doppler effect equation, we can calculate the initial observed frequency as:

f' = f(1 - (v_o/v))

We are given;

f = 46.2 kHz = 46200 Hz

v_o = 21.8 m/s

v is speed of sound = 343 m/s

Thus;

f' = 46200(1 - (21/343))

f' = 43371.4285 Hz

In the second stage, we see that the bat is now a stationary observer while the object is now the moving source;

Thus, from doppler effect again but this time with the source going away from the obsever, the new observed frequency is;

f" = f'/(1 + (v_o/v))

f" = 43371.4285/(1 + (21.8/343))

f" = 40779.61 Hz

4 0
3 years ago
How do I solve for the maximum height of someone who jumps straight up leaving the ground at 6.0 m/s.
Luda [366]

Answer: 3.0 m/s

Explanation:

5 0
3 years ago
A fuel pump sends gasoline from a car's fuel tank to the engine at a rate of 6.55x10-2 kg/s. The density of the gasoline is 740
klasskru [66]

Answer:

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6.55 x 10^-2 / 740 = pi * (2.67 x 10^-3)^2 * speed

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6 0
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Georgia [21]

Answer:

The answer is below

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We are to check if the statement is true of false. If it is false, we correct the statement.

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Acceleration = \frac{24-0}{4-0}=6\ m/s^2

This is correct.

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Sloan [31]

Answer:

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