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4 years ago

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An Earth satellite moves in a circular orbit 561 km above Earth's surface with a period of 95.68 min. What are (a) the speed and

(b) the magnitude of the centripetal acceleration of the satellite?

1 answer:

crimeas [40]4 years ago

3 0

Explanation:

It is given that,

Radius of earth, r = 6371 km

An earth satellite moves in a circular orbit above the Earth's surface, d = 561 km

So, radius of satellite, R = 6371 km + 561 km = 6932 × 10³ m

Time taken, t = 95.68 min = 5740.8 sec

(a) Speed of the satellite is given by :

$v=\dfrac{d}{t}$

d = distance covered

For circular path, d = 2πR

$v=\dfrac{2\pi \times 6932\times 10^3\ m}{5740.8\ sec}$

v = 7586.92 m/s

(b) Centripetal acceleration is given by :

$a=\dfrac{v^2}{R}$

$a=\dfrac{(7586.92\ m/s)^2}{6932\times 10^3\ m}$

$a=8.3\ m/s^2$

Hence, this is the required solution.

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Scrat [10]

Answer:

The current induced in the loop is $\dfrac{\pi a^2 B\omega}{4R}\ \sin(\omega t)$ .

Explanation:

Given that,

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3 years ago

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BigorU [14]

Answer:

1.6 x 10^{-19} Coulombs

Explanation:

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In Chemistry, electrons can be defined as subatomic particles that are negatively charged and as such has a magnitude of -1.

The minimum charge on any object such as an electron cannot be less than 1.6 x 10^{-19} Coulombs and it's usually referred to as the fundamental unit of charge.

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3 years ago

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kipiarov [429]

Answer:

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4 years ago

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photoshop1234 [79]

Answer:

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3 years ago

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And the negative sign tells that the direction of the acceleration is opposite to that of the velocity.

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4 years ago