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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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2 years ago
A soft tennis ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.10 m. (a) Calculate its ve
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Answer:

(a)   v = 5.42m/s

(b)   vo = 4.64m/s

(c)   a = 2874.28m/s^2

(d)   Δy = 5.11*10^-3m

Explanation:

(a) The velocity of the ball before it hits the floor is given by:

v=\sqrt{2gh}        (1)

g: gravitational acceleration = 9.8m/s^2

h: height where the ball falls down = 1.50m

v=\sqrt{2(9.8m/s^2)(1.50m)}=5.42\frac{m}{s}

The speed of the ball is 5.42m/s

(b) To calculate the velocity of the ball, after it leaves the floor, you use the information of the maximum height reached by the ball after it leaves the floor.

You use the following formula:

h_{max}=\frac{v_o^2}{2g}       (2)

vo: velocity of the ball where it starts its motion upward

You solve for vo and replace the values of the parameters:

v_o=\sqrt{2gh_{max}}=\sqrt{2(9.8m/s^2)(1.10m)}=4.64\frac{m}{s}

The velocity of the ball is 4.64m/s

(c) The acceleration is given by:

a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-(-5.42m)/s}{3.50*10^{-3}s}=2874.285\frac{m}{s^2}

a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-5.42m/s}{3.50*10^{-3}s}=-222.85\frac{m}{s^2}

The acceleration of the ball is 2874.28/s^2

(d) The compression of the ball is:

\Deta y=\frac{v^2}{2(a)}=\frac{(5.42m/s)^2}{2(2874.28m/s^2)}=5.11*10^{-3}m

THe compression of the ball when it strikes the floor is 5.11*10^-3m

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