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Yanka [14]
3 years ago
8

Consider three identical metal spheres, A, B, and C. Sphere A carries a charge of +6q. Sphere B carries a charge of -q. Sphere C

carries no net charge. Spheres A and B are touched together and then separated. Sphere C is then touched to sphere A and separated from it. Lastly, sphere C is touched to sphere B and separated from it. (a) What is the ratio of the final charge on sphere C to q? What is the ratio of the final total charge on the three spheres to q (b) before they are allowed to touch each other and (c) after they have touched?
Physics
1 answer:
elena-14-01-66 [18.8K]3 years ago
6 0

Answer:

Part a)

Final charge on C : q = 1.875

Part b)

Ratio for A = 6 : 1.25

Ratio for B = -1 : 1.875

Ratio for C = 0

Explanation:

When two identical metal sphere are connected together then the charge on them will get equally divided on both after connecting them by conducting wire

So here we have

q_A = + 6q

q_B = -q

q_c = 0

Step 1: We connected A and B and then separate them

so we have

q_A' = q_B' = 2.5q

Step 2: We connected A and C and then separate them

so we have

q_A'' = q_c' = 1.25q

Step 3: We connected B and C and then separate them

so we have

q_c'' = q_b'' = 1.875q

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Answer:

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Explanation:

The change in length on an object due to rise in temperature is given by the following equation of linear thermal expansion:

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Therefore,

ΔL = (11 x 10⁻⁶ °C⁻¹)(1000 m)(60°C)

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Answer:

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The value of 1.0004 to the power 1 by 2 using Binomial approximation is​
IgorLugansk [536]

Given:

The given value is (1.0004)^{\frac{1}{2}}.

To find:

The value of the given expression by using the Binomial approximation.

Explanation:

We have,

(1.0004)^{\frac{1}{2}}

It can be written as:

(1.0004)^{\frac{1}{2}}=(1+0.0004)^{\frac{1}{2}}

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