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4 years ago

At height h above the surface of Earth, the gravitational acceleration is What is h? Note: The radius of Earth is 6380 km.

1 answer:

4 0

The acceleration of gravity is inversely proportional to
the square of the distance from Earth's center.

The acceleration of gravity is 9.8 m/s² on the Earth's surface ...
6380 km from the center.

If the acceleration of gravity at 'h' is 4.9 m/s² ... 1/2 of what it is
on the surface, then the distance from the center is

(6380 x √2) = 9,023 km (rounded) ,

and 'h' is the distance above the surface

= (9,023 - 6,380) = 2,643 km (rounded) .

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3 years ago

Test yourself: 1

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Answer:

F = 1500 [N]

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4 years ago

What are materials that doesn't not use electromagnetism

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Answer:

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3 years ago

A ball is whirled on the end of a string in a horizontal circle of radius R at constant speed v. Which way(s) can increase the c

Charra [1.4K]

Explanation:

When an object moves in a circular path, due to the change in its velocity, the object possess centripetal acceleration. The formula for the centripetal acceleration is given by :

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

Where

v is the speed of the object

R is the radius of the circle path

Option (1) : Keeping the speed fixed and decreasing the radius by a factor of 9

$a=\dfrac{v^2}{R/9}$

$a=\dfrac{9v^2}{R}$

The centripetal acceleration of the ball by a factor of 9.

Option (2) : Keeping the radius fixed and increasing the speed by a factor of 3

$a=\dfrac{(3v)^2}{R}$

$a=\dfrac{9v^2}{R}$

The centripetal acceleration increases by a factor of 9.

Option (3) : Decreasing both the radius and the speed by a factor of 9.

$a=\dfrac{(v/9)^2}{R/9}$

$a=\dfrac{(v)^2}{9R}$

The centripetal acceleration decreases by a factor of 9.

Option (4) : Keeping the radius fixed and increasing the speed by a factor of 9

$a=\dfrac{(3v)^2}{R}$

$a=\dfrac{9v^2}{R}$

The centripetal acceleration increases by a factor of 9.

Option (5) : Increasing both the radius and the speed by a factor of 9

$a=\dfrac{(9v)^2}{9R}$

$a=\dfrac{9v^2}{R}$

The centripetal acceleration increases by a factor of 9.

Option (6) : Keeping the speed fixed and increasing the radius by a factor of 9

$a=\dfrac{(v)^2}{9R}$

$a=\dfrac{9v^2}{R}$

The centripetal acceleration increases by a factor of 9.

So, as the radius of the circle decreases, its centripetal acceleration increase. Also, if the speed of the object increases, its centripetal acceleration increase. Hence, this is the required solution.

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3 years ago