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2 years ago

When a satellite is in orbit around the earth the force of gravity on the satellite-

Physics

1 answer:

mojhsa [17]2 years ago

5 0

Answer:

Is always towards the center of the Earth

Explanation:

As a satellite moves around the Earth in a circular orbit, the direction of the force of gravity is always towards the center of the Earth. At an altitude of 100 km, you would be so high that you would see black sky and stars if you looked upwards.

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In the fishbowl, the glass, water, rocks, and plastic plants are in thermal equilibrium. This situation means the temperature of

Assoli18 [71]

To cold which it throws off the equilibrium of the other things and sticks to it longer

7 0

1 year ago

6. (a) Suppose the earth is revolving round the sun in a circular orbit of radius one b astronomical unit (1.5% 10 km). Find the

kkurt [141]

Answer:

tough

ques

Explanation:

5 0

1 year ago

One baked potato provides an average of 30 mg of vitamin C. If 70 potatoes weigh 20 lb., how many milligrams of vitamin C are pr

JulsSmile [24]

Answer:

105 mg

Explanation:

Given that:

1 baked potato provides 30 mg of vitamin C.

So,

70 baked potatoes provide $30\times 70$ mg of vitamin C

Also,

70 potatoes = 20 lb

So,

20 lb potatoes provide $30\times 70$ mg of vitamin C

Thus,

1 lb potatoes provide $\frac {30\times 70}{20}$ mg of vitamin C

<u>Thus, 105 mg of Vitamin C are provided per pound of the potatoes.</u>

5 0

2 years ago

A spherical asteroid of average density would have a mass of 8.7×1013kg if its radius were 2.0 km.A)If you and your spacesuit ha

WITCHER [35]

A) 0.189 N

The weight of the person on the asteroid is equal to the gravitational force exerted by the asteroid on the person, at a location on the surface of the asteroid:

$F=\frac{GMm}{R^2}$

where

G is the gravitational constant

8.7×10^13 kg is the mass of the asteroid

m = 130 kg is the mass of the man

R = 2.0 km = 2000 m is the radius of the asteroid

Substituting into the equation, we find

$F=\frac{(6.67\cdot 10^{-11})(8.7\cdot 10^{13} kg)(130 kg)}{(2000 m)^2}0.189 N=$

B) 2.41 m/s

In order to orbit just above the surface of the asteroid (r=R), the centripetal force that keeps the astronaut in orbit must be equal to the gravitational force acting on the astronaut:

$\frac{GMm}{R^2}=\frac{mv^2}{R}$