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valentina_108 [34]
2 years ago
14

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?
A) It deflects more than it did at the end of the video.
B) It does not deflect at all.
C) It deflects the same amount as at end of the video.
D) It deflects less than it did at the end of the video.
Physics
1 answer:
horrorfan [7]2 years ago
8 0

Note: Although the video is not provided in this question, it is not needed to answer the question.

Answer:

B) It does not deflect at all

Explanation:

Since both half shells contain opposite charges, the two shells become electrically neutral when they are brought together and the electroscope discharges. On separating the two half shells again, the needle does not deflect because the half shells have now lost their charges to become neutral.

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A wheel, starting from rest, rotates with a constant angular acceleration of 1.80 rad/s^2. During a certain 7.00 s interval, it
lora16 [44]

Answer:

a) 1.3 rad/s

b) 0.722 s

Explanation:

Given

Initial velocity, ω = 0 rad/s

Angular acceleration of the wheel, α = 1.8 rad/s²

using equations of angular motion, we have

θ2 - θ1 = ω(0)[t2 - t1] + 1/2α(t2 - t1)²

where

θ2 - θ1 = 53.2 rad

t2 - t1 = 7s

substituting these in the equation, we have

θ2 - θ1 = ω(0)[t2 - t1] + 1/2α(t2 - t1)²

53.2 =ω(0) * 7 + 1/2 * 1.8 * 7²

53.2 = 7.ω(0) + 1/2 * 1.8 * 49

53.2 = 7.ω(0) + 44.1

7.ω(0) = 53.2 - 44.1

ω(0) = 9.1 / 7

ω(0) = 1.3 rad/s

Using another of the equations of angular motion, we have

ω(0) = ω(i) + α*t1

1.3 = 0 + 1.8 * t1

1.3 = 1.8 * t1

t1 = 1.3/1.8

t1 = 0.722 s

5 0
3 years ago
You push your friend, whose mass is 54kg, down a hill so she can go sledding. Her acceleration is 3m/s2. Calculate the amount of
Alecsey [184]

Answer:

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3 0
2 years ago
A charge of 8.5 × 10–6 C is in an electric field that has a strength of 3.2 × 105 N/C. What is the electric force acting on the
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2.72 N

Explanation:

Step 1:

From the basic formula in electrostatics

F = E * q

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Step 2:

From the given question

q= 8.5*10^{-6} C

E = 3.2 * 10^{5} N/C

F = 8.5 * 10^{-6} * 3.2 * 10^{5} = 2.72 N

8 0
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From Carnot's theorem, for any engine working between these two temperatures:


efficiency <= (1-tc/th) * 100


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For a, th = 900k, efficiency = (1-300/900) = 70%

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Hence in case of a and b, efficiency claimed is lesser than efficiency calculated, which is valid case and in case of c, however efficiency claimed is greater which is invalid. 

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Irina-Kira [14]

Answer:

50*

Explanation:

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