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kati45 [8]
3 years ago
8

Satellites in near-earth orbit experience a very slight drag due to the extremely thin upper atmosphere. These satellites slowly

but surely spiral inward, where they finally burn up as they reach the thicker lower levels of the atmosphere. The radius decreases so slowly that you can consider the satellite to have a circular orbit at all times. As a satellite spirals inward, does it speed up, slow down, or maintain the same speed?
Physics
1 answer:
erica [24]3 years ago
6 0
According to my research, a satellite in a circular Earth orbit is subject to a very tiny constant friction force, due to the atmosphere. As it spirals inward, it slowly decreases its orbital radius.
So as long as a satellite moves in a circular orbit, it's velocity is inversely proportionate to the square root of the orbital radius-it speeds up as it spirals inward. I hope my answer proves of some help to you.
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Ira Lisetskai [31]

Answer:

If we want to reach the planet PSR B1620-26 b, explain why we will need to make some big “wrinkle in time” discoveries or find ways to live much, much longer?

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If we want to reach the planet PSR B1620-26 b, explain why we will need to make some big “wrinkle in time” discoveries or find ways to live much, much longer?

Explanation:

If we want to reach the planet PSR B1620-26 b, explain why we will need to make some big “wrinkle in time” discoveries or find ways to live much, much longer?

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If we want to reach the planet PSR B1620-26 b, explain why we will need to make some big “wrinkle in time” discoveries or find ways to live much, much longer?

5 0
2 years ago
Read 2 more answers
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dimaraw [331]
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3 0
3 years ago
A harmonic wave on a string with a mass per unit length of 0.050 kg/m and a tension of 60 N has an amplitude of 5.0 cm. Each sec
Dennis_Churaev [7]

Answer:

Power of the string wave will be equal to 5.464 watt

Explanation:

We have given mass per unit length is 0.050 kg/m

Tension in the string T = 60 N

Amplitude of the wave A = 5 cm = 0.05 m

Frequency f = 8 Hz

So angular frequency \omega =2\pi f=2\times 3.14\times 8=50.24rad/sec

Velocity of the string wave is equal to v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{60}{0.050}}=34.641m/sec

Power of wave propagation is equal to P=\frac{1}{2}\mu \omega ^2vA^2=\frac{1}{2}\times 0.050\times 50.24^2\times 34.641\times 0.05^2=5.464watt

So power of the wave will be equal to 5.464 watt

6 0
3 years ago
An ant does 1 N-m of work in dragging a 0.0020-N grain of sugar. How far does the ant drag the sugar.
algol13
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Good luck!
8 0
3 years ago
If he leaves the ramp with a speed of 35.0 m/s and has a speed of 33.0 m/s at the top of his trajectory, determine his maximum h
nadezda [96]

Answer:

H = 6.93 m

Explanation:

given data

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solution

we get here maximum height so first we get vertical component here that is express as

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put here value

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put here value

H = \frac{11.66^2}{2\times 9.8}

H = 6.93 m

7 0
3 years ago
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