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Kamila [148]
3 years ago
10

A soccer player takes a corner kick, lofting a stationary ball 33.0° above the horizon at 15.0 m/s. If the soccer ball has a mas

s of 0.425 kg and the player's foot is in contact with it for 5.10 ✕ 10−2 s, find the x- and y-components of the soccer ball's change in momentum and the magnitude of the average force exerted by the player's foot on the ball.
PART A: the x- and y-components of the soccer ball's change in momentum (in kg · m/s)

Δpx ?????????? = kg · m/s

Δpy ?????????? = kg · m/s
Physics
1 answer:
Alexxandr [17]3 years ago
3 0

Explanation:

It is given that,

Mass of the soccer ball, m = 0.425 kg

Speed of the ball, u = 15 m/s

Angle with horizontal, \theta=33^{\circ}

Time for which the player's foot is in contact with it, \Delta t = 5.1\times 10^{-2}\ s

Part A,

The x component of the soccer ball's change in momentum is given by :

\Delta p_x=mv\ cos\theta

\Delta p_x=0.425\times 15\ cos(33)

p_x=5.34\ kg-m/s

The y component of the soccer ball's change in momentum is given by :

\Delta p_y=mv\ sin\theta

\Delta p_y=0.425\times 15\ sin(33)

p_y=3.47\ kg-m/s

Hence, this is the required solution.

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1.2 m long spring with a spring force of 200.0 N/m, 12.0 kg mass is attached to it, find the length
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From Hooke's law, the length of the spring after extension and after the mass is attached is 1.788 meters. Option B is the answer.

<h3>HOOKE'S LAW</h3>

Hooke's law state that in an elastic material, the force applied is directly proportional to the extension provided that the elastic limit is not exceeded.

Given that a 12.0 kg mass is attached to 1.2 m long spring with a spring constant of 200.0 N/m.

The given parameters are:

  • mass m = 12kg
  • Initial length L_{1} = 1.2 m
  • Spring constant K = 200 N/m
  • Extension e = ?

According to Hooke's law

F = Ke

But F = mg

mg = Ke

Substitute all the parameters into the formula to get extension e

12 x 9.8 = 200e

e = 117.6 / 200

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The length of the spring after extension and after the mass is attached will be

L_{2} = L_{1} + e

L_{2} = 1.2 + 0.588

L_{2} = 1.788 m

Therefore, the correct answer is option B because the length of the spring after extension and after the mass is attached is 1.788 meters.

Learn more about Hooke's law here: brainly.com/question/12253978

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