If the gravitational field strength at the top of Mount Everest is 9.772 N/kg, approximately how tall (in feet) is

Physics

1 answer:

Delicious77 [7]3 years ago

6 0

Answer:

The height of the Everest mountain is, x = 8514.087 m

Explanation:

Given data,

The gravitational field strength at the top of mount Everest, gₓ = 9.772 m/s²

The formula for gravitational field strength is,

Where, x is the height from the surface of the Earth

Therefore,

(R+x)² = GM/gₓ

x = √(GM/gₓ) - R

Substituting the values,

x = √(6.67408 x 10⁻¹¹ X 5.972 x 10²⁴ / 9.772) - 6.378 x 10⁶

x = 8514.087 m

Therefore, the height of the Everest mountain is, x = 8514.087 m

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Mass m moves to the right with speed =v along a frictionless horizontal surface and crashes into an equal mass m initially at re

Amiraneli [1.4K]

After the collision the magnitude of the momentum of the system is Mv

Given:

mass of 1st object = M

speed of 1st object = v

mass of 2nd object = M

speed of 2nd object = 0

To Find:

magnitude of the momentum after collision

Solution: Product of the mass of a particle and its velocity. Momentum is a vector quantity; i.e., it has both magnitude and direction. Isaac Newton's second law of motion states that the time rate of change of momentum is equal to the force acting on the particle.

Applying conservation of linear momentum

Mv + M(0) = 2MV

Mv = 2MV

V = v/2

So, after collision momentum is

p = 2MV = 2xMxv/2 = Mv

So, after collision momentum is Mv

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2 years ago