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castortr0y [4]
3 years ago
15

A 91.5 kg football player running east at 2.73 m/s tackles a 63.5 kg player running east at 3.09 m/s. what is their velocity aft

erward? PLEASE HELP
Physics
1 answer:
matrenka [14]3 years ago
3 0

Answer:

The velocity of the players will be <u>2.88 m/s</u> in the <u>east</u> direction.

Explanation:

Let 'v' be the velocity of the players after collision.

Consider the east direction as positive direction.

Given:

Mass of the first player  is, m_1 = 91.5 kg  

Initial velocity of the first player  is, u_1 = 2.73 m/s  

Mass of the second player  is, m_2 = 63.5 kg  

Initial velocity of the second player is, u_2 = 3.09 m/s  

In order to solve this problem we use the law of conservation of momentum which says that the total momentum must be conserved before and after the collision. So we can write:

p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1 + m_2)v

Solving for v, we get:

v = \frac{m_1 u_1+m_2 u_2}{m_1+m_2}=\frac{(91.5)(2.73)+(63.5)(3.09)}{91.5+63.5}=2.88\ m/s

Therefore, their velocity after the collision is 2.88 m/s.

The sign of the velocity after collision is positive. So, the players will move in the east direction only after collision.

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<u>Answer</u>: The mass of the object is 25kg.

The given question deals with Newton's second law of motion and its applications.

<u>Explanation:</u> Given force, F=500N

                                 acceleration, a=20 m/s^{2}

  From Newton's 2nd law of motion , we have

                             F=ma where m=mass of the object

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<u> </u><u>Reference Link: </u>brainly.com/question/1141170

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Answer:

Explanation:

Given

charge of first body q_1=2.5\ mu C

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third Particle which charge +q must be placed left of 2.5\mu C because it will repel the q charge while -3.5\mu C will attract it

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F_{1q}+F_{2q}=0

\frac{kq(2.5)}{x^2}+\frac{kq(-3.5)}{(0.6+x)^2}=0

\frac{kq(2.5)}{x^2}=\frac{kq(3.5)}{(0.6+x)^2}

\frac{0.6+x}{x}=(\frac{3.5}{2.5})^{0.5}

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