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posledela
3 years ago
11

We can use the textbook results for head-on elastic collisions to analyze the recoil of the Earth when a ball bounces off a wall

embedded in the Earth. Suppose a professional baseball pitcher hurls a baseball (m = 155 grams = 0.155 kg) with a speed of 103 miles per hour (vball = 45.3 m/s) at a wall, and the ball bounces back with little loss of kinetic energy.(a) What is the recoil speed of the Earth (M = 6multiply1024 kg)?

Physics
2 answers:
hjlf3 years ago
4 0

Answer:

Recoil velocity; v2 = 1.17 x 10^(-24) m/s

Explanation:

We are given;

Mass of baseball; m1 = 0.155 kg

Velocity of baseball; v1 = 45.3 m/s

Mass of earth; m2 = 6 x 10^(24) kg

Lets the recoil velocity be v2.

From conservation of momentum, initial momentum is equal to final momentum.

Thus,

m1•v1 = m2v2

Thus, v2 = m1•v1/m2

Plugging in the relevant values to get ;

v2 = 0.155 x 45.3/(6 x 10^(24))

v2 = 1.17 x 10^(-24) m/s

Nataly [62]3 years ago
3 0

Answer:

The explaination and attachment better portrays the answer. kindly check explaination.

Explanation:

To be able to solve for the kinetic energy and the recoil speed. Let have an understanding of those words.

Kinetic energy is the energy of motion. If an object is moving, it is said to have kinetic energy.

Please kindly check attachment for the detailed answer.

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The angular speed of the device is 1.03 rad/s.

<h3>What is the conservation of angular momentum?</h3>

A spinning system's ability to conserve angular momentum ensures that its spin will not change until it is subjected to an external torque; to put it another way, the rotation's speed will not change as long as the net torque is zero.

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The moment of inertia of the system is given by

I = 2(M₁R₁² + M₂R₂²)+ mR₁²

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Here,  = Iω

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1.692 = 1.6292(ω)

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To know more about the conservation of angular momentum, visit:

brainly.com/question/1597483

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