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AlexFokin [52]
3 years ago
15

A satellite is in orbit 36000km above the surface of the earth. Its angular velocity is 7.27*10^-5 rad/s. What is the velocity o

f the satellite? (The radius of the earth is 6400km.)
Physics
1 answer:
bagirrra123 [75]3 years ago
6 0

Answer:

v = 3.08 km/s

Explanation:

Given that,

The angular velocity of the satellite = \omega=7.27\times 10^{-5} rad/s

A satellite is in orbit 36000km above the surface of the earth.

The radius of the earth is 6400 km

Let v is the velocity of the satellite. It can be calculated as :

v=r\omega\\\\v=(36000\times 10^3+6400\times 10^3)\times 7.27\times 10^{-5}\\\\v=3082.48\ m/s\\\\v=3.08\ km/s

So, the velocity of the satellite is 3.08 km/s.

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A proton is moved from the negative to the positive plate of a parallel-plate arrangement. The plates are 1.50 cm apart, and the
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(a) The proton’s potential energy change is 3.6 x 10⁻¹⁸ J.

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U = (1.6 x 10⁻¹⁹)(1500 x 1.5 x 10⁻²)

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The potential difference between the negative plate and a point midway between the plates is 11.25 V.

The speed of the proton just before it hits the negative plate is 6.57 x 10⁴ m/s.

Learn more about potential difference here: brainly.com/question/24142403

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