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balu736 [363]
3 years ago
14

A person is placed in a large, hollow, metallic sphere that is insulated from ground. (a) If a large charge is placed on the sph

ere, will the person be harmed upon touching the inside of the sphere? Yes No Correct: Your answer is correct. (b) Explain what will happen if the person also has an initial charge whose sign is opposite that of the charge on the sphere.
Physics
1 answer:
stiv31 [10]3 years ago
7 0

Answer:

Explanation:

It is given that the sphere is insulated from ground and a large charge is placed on the sphere. The charge on the hollow sphere will always remain on the outer surface of the sphere and there will be no charge on the inner surface of the sphere.

If a person touches the inner surface of the sphere then he will not be harmed as there is no charge on the inner surface of the sphere.

If a person carries the charge of the opposite sign of the same magnitude then the sphere and person get neutralized upon touching the sphere.

If a person does not touches the sphere then the charge on the outer surface will be zero and there will be a positive charge on the inner surface of the sphere                        

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2. A small car has a mass of 890 kg. What is its weight? (Hint: remember the definition of weight).
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or

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The Apollo Lunar Module was used to make the transition from the spacecraft to the moon's surface and back. Consider a similar m
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Answer:

Explanation:

a. Landing height is

H=1.3m

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u=1.3m/s

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v=?

The acceleration due to gravity is 0.4 times that of the one on earth,

Then, g on earth is approximately 9.81m/s²

Then, g on Mars is

g=0.4×9.81=3.924m/s²

Then using equation of motion for a free fall body

v²=u²+2gH

v²=1.3²+2×3.924×1.3

v²=1.69+10.2024

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v=√11.8924

v=3.45m/s

The impact velocity of the spacecraft is 3.45m/s

b. For a lunar module, the safe velocity landing is 3m/s

v=3m/s.

Given that the initial velocity is 1.2m/s²

We already know acceleration due to gravity on Mars is g=3.924m/s²

The we need to know the maximum height to have a safe velocity of 3m/s

Then using equation of motion

v²=u²+2gH

3²=1.2²+2×3.924H

9=1.44+7.848H

9-1.44=7.848H

7.56=7.848H

H=7.56/7.848

H=0.963m

The the maximum safe landing height to obtain a final landing velocity of 3m/s is 0.963m

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Answer:

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Lead=4

Bismuth=5

Potassium=1

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