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solong [7]
3 years ago
11

What will happen to a spacecraft that is moving far out in space, and that is outside Earth’s atmosphere and the pull of gravity

?
A. It will continue to move at the same speed and in the same direction.
B. It will eventually slow down and fall to Earth.
C. It will continue to move faster and faster away from Earth.
D. It will burn up due to friction.
Physics
1 answer:
Akimi4 [234]3 years ago
4 0
I think the answer is D
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Difference between freefall and weightlessness.​
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Answer:

Differences between freefall and weightlessness are as follows:

<h3><u>Freefall</u></h3>
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<h3><u>Weightlessness</u></h3>
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The moment of water from ocean through the atmosphere and back
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There is always water in the atmosphere. Clouds are, of course, the most visible manifestation of atmospheric water, but even clear air contains water—water in particles that are too small to be seen. One estimate of the volume of water in the atmosphere at any one time is about 3,100 cubic miles (mi3) or 12,900 cubic kilometers (km3). That may sound like a lot, but it is only about 0.001 percent of the total Earth's water volume of about 332,500,000 mi3 (1,385,000,000 km3), If all of the water in the atmosphere rained down at once, it would only cover the globe to a depth of 2.5 centimeters, about 1 inch.

6 0
3 years ago
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A circular loop of wire with radius 2.00 cm and resistance 0.600 Ω is in a region of a spatially uniform magnetic field B⃗ that
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Answer:

Incomplete questions

Let assume we are asked to find

Calculate the induced emf in the coil at any time, let say t=2

And induced current

Explanation:

Flux is given as

Φ=NAB

Where

N is number of turn, N=1

A=area=πr²

Since r=2cm=0.02

A=π(0.02)²=0.001257m²

B=magnetic field

B(t)=Bo•e−t/τ,

Where Bo=3T

τ=0.5s

B(t)=3e(−t/0.5)

B(t)=3exp(-2t)

Therefore

Φ=NAB

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Φ=0.00377exp(-2t)

Now,

Induce emf is given as

E= - dΦ/dt

E= - 0.00377×-2 exp(-2t)

E=0.00754exp(-2t)

At t=2

E=0.00754exp(-4)

E=0.000138V

E=0.138mV

b. Induce current

From ohms laws

V=iR

Given that R=0.6Ω

i=V/R

i=0.000138/0.6

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i=0.23mA

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