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Paladinen [302]

4 years ago

9

The equation for the speed of a satellite in a circular orbit around the earth depends on mass. Which mass?

1 answer:

katovenus [111]4 years ago

6 0

<h3><u>Question: </u></h3>

The equation for the speed of a satellite in a circular orbit around the Earth depends on mass. Which mass?

a. The mass of the sun

b. The mass of the satellite

c. The mass of the Earth

<h3><u>Answer:</u></h3>

The equation for the speed of a satellite orbiting in a circular path around the earth depends upon the mass of Earth.

Option c

<h3><u> Explanation: </u></h3>

Any particular body performing circular motion has a centripetal force in picture. In this case of a satellite revolving in a circular orbit around the earth, the necessary centripetal force is provided by the gravitational force between the satellite and earth. Hence $F_{G} = F_{C}$ .

Gravitational force between Earth and Satellite: $F_{G} = \frac{G \times M_e \times M_s}{R^2}$

Centripetal force of Satellite : $F_C = \frac{M_s \times V^2}{R}$

Where G = Gravitational Constant

$M_e$ = Mass of Earth

$M_s$ = Mass of satellite

R= Radius of satellite’s circular orbit

V = Speed of satellite

Equating $F_G = F_C$ , we get

Speed of Satellite $V =\frac{\sqrt{G \times M_e}}{R}$

Thus the speed of satellite depends only on the mass of Earth.

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3 years ago

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A 60-kg rollerblader rolls 10 m down a 30? incline. When she reaches the level floor at the bottom, she applies the brakes. The

VARVARA [1.3K]

Answer:

s = 20 m

Explanation:

given,

mass of the roller blader = 60 Kg

length = 10 m

inclines at = 30°

coefficient of friction = 0.25

using conservation of energy

$\dfrac{1}{2}mu^2 = m g d sin \theta$

$u^2 = 2 g d sin30^0$

$u= \sqrt{2\times 9.8 \times 10 sin30^0}$

u = 9.89 m/s

Using second law of motion

ma =μ mg

a = μ g

a = 0.25 x 9.8

a = 2.45 m/s²

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3 0

3 years ago

On a balanced seesaw, a boy three times as heavy as his partner sits

slega [8]

Answer:

1/3 the distance from the fulcrum

Explanation:

On a balanced seesaw, the torques around the fulcrum calculated on one side and on another side must be equal. This means that:

$W_1 d_1 = W_2 d_2$

where

W1 is the weight of the boy

d1 is its distance from the fulcrum

W2 is the weight of his partner

d2 is the distance of the partner from the fulcrum

In this problem, we know that the boy is three times as heavy as his partner, so

$W_1 = 3 W_2$

If we substitute this into the equation, we find:

$(3 W_2) d_1 = W_2 d_2$

and by simplifying:

$3 d_1 = d_2\\d_1 = \frac{1}{3}d_2$

which means that the boy sits at 1/3 the distance from the fulcrum.

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3 years ago

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3 years ago

A motorist travels at 270 km in p hours at a certain average speed. an expression for the distance that this motorist travelled

wel

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Average speed = 270 km / p hours
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5 0

3 years ago