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grigory [225]
3 years ago
6

An Earth satellite is orbiting at a distance from the Earth’s surface equal to one Earth radius (4 000 miles). At this location,

the acceleration due to gravity is what factor times the value of gat the Earth’s surface?
A. There is no acceleration since the satellite is in orbit.
B. 2
C. 1/2
D. 1/4
Physics
1 answer:
trasher [3.6K]3 years ago
4 0

Answer:

The acceleration due to gravity is \dfrac{1}{4} times the value of g at the Earth’s surface.

(D) is correct option.

Explanation:

Given that,

Radius = 4000 miles

We need to calculate the gravitational force at surface

Gravitational force on the mass m on the surface of the earth

At r = R

F=mg

mg=\dfrac{GmM}{R^2}....(I)

We need to calculate the gravitational force at height

Gravitational force on a mass m from the center of the earth,

At r = R + R = 2 R

F'=mg'

mg'=\dfrac{GmM}{4R^2}....(II)

Dividing equation (II) by equation (I)

\dfrac{mg'}{mg}=\dfrac{\dfrac{GmM}{4R^2}}{\dfrac{GmM}{R^2}}

\dfrac{g'}{g}=\dfrac{1}{4}

Hence, The acceleration due to gravity is \dfrac{1}{4} times the value of g at the Earth’s surface.

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Being able to see a square in the middle of the image despite not having lines to form a square represents the Gestalt principle
vredina [299]

Answer: The correct answer for the blank is -

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Read 2 more answers
A commuter backs her car out of her garage with an acceleration of 1.40 m/s2.A) How long does it take her to reach a speed of 2.
AURORKA [14]

Answer:

(A) 1.43secs

(B) -2.50m/s^2

Explanation:

A commuter backs her car out of her garage with an acceleration of 1.40m/s^2

(A) When the speed is 2.00m/s then, the time can be calculated as follows

t= Vf-Vo/a

The values given are a= 1.40m/s^2 , Vf= 2.00m/s, Vo= 0

= 2.00-0/1.40

= 2.00/1.40

= 1.43secs

(B) The deceleration when the time is 0.800secs can be calculated as follows

a= Vf-Vo/t

= 0-2.00/0.800

= -2.00/0.800

= -2.50m/s^2

4 0
3 years ago
What is the meaning of the reference point in electric potential?.
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Answer:

<h3><u>ELECTRIC POTENTIAL</u></h3>

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4 0
2 years ago
When 10 N force applied at 30 degrees to the end of a 20 cm handle of a wrench, it was just able to loosen the nut. What magnitu
fredd [130]

Answer:

<h2>5N</h2>

Explanation:

To get the magnitude of the force would require to just loosen the nut, if the force apply perpendicularly at the end of the handle, we will have to resolve the force perpendicular to the wrench. Torque is the turning effect of a body or force about a point. It is similar to moments.

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Note that the force must be perpendicular to the wrench. On resolving the force perpendicularly to the wrench, we will have to resolve the force to the vertical.

Fy = Fsinθ

Fy = 10sin30°

Fy = 10 * 0.5

Fy = 5N

<em>Torque = Fy * r</em>

<em>Given Fy = 5N and r = 20cm = 0.2m</em>

<em>Torque = 5 * 0.2</em>

<em>Torque = 1Nm</em>

<em />

<em>Hence the magnitude of the force would require to just loosen the nut, if the force apply perpendicularly at the end of the handle is 5N</em>

3 0
3 years ago
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