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guapka [62]
3 years ago
14

The mass of the Earth is 5.98 × 1024 kg . A 11 kg bowling ball initially at rest is dropped from a height of 2.63 m. The acceler

ation of gravity is 9.8 m/s 2 . What is the speed of the Earth coming up to meet the ball just before the ball hits the ground? Answer in units of m/s.
Physics
1 answer:
jekas [21]3 years ago
3 0

Answer:

v = 7.18_m/s

Explanation:

Velocity of the earth towards the ball is = velocity of the ball moving towards earth

For object in free fall, we have

Where

v = final velocity

u = initial velocity

g = acceleration due to gravity

t = time

S = height of ball above ground

v^2 = u^2 - 2×g×(-S)

= 0 + 2×9.8×2.63 = 51.55_m^2/s^2

Velocity of the ball just before it hits the ground is

v = 7.18_m/s

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3 years ago
Christina drives his moped 7 kilometers North. She stops for lunch and then drives 5 kilometers East. What distance did she cove
Mashutka [201]

Answer:

She covers the distance is 12 km.

The magnitude of displacement is 8.6 km.

The direction of her displacement is north east.

Explanation:

Given that,

Christina drives his moped 7 kilometers North and stop for lunch and then drive 5 km east.

We need to calculate the total distance

Using formula of distance

d=d_{1}+d_{2}

Put the value into the formula

d=7+5

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Hence, She covers the distance is 12 km.

The magnitude of displacement is 8.6 km.

The direction of her displacement is north east.

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3 years ago
A 0.14-MIN baseball is dropped from rest. It has a momentum of 0.90 kg⋅m/skg⋅m/s just before it lands on the ground.
nikitadnepr [17]

The time spent in the air by the ball at the given momentum is 6.43 s.

The given parameters;

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The impulse experienced by the ball is calculated as follows;

Ft = \Delta P

where;

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\Delta P is change in momentum

The time of motion of the ball is calculated as follows;

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Thus, the time spent in the air by the ball at the given momentum is 6.43 s.

Learn more here:brainly.com/question/13468390

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I am going to say

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