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tia_tia [17]
2 years ago
7

A 50 kg pitcher throws a baseball with a mass of 0.15 kg. If the ball is thrown with a positive velocity of 35 m/s and there is

no net force on the system, what is the velocity of the pitcher?
Physics
1 answer:
Andreas93 [3]2 years ago
8 0

The velocity of the pitcher is <u>0.105 m/s</u> in a direction opposite to the velocity of the ball.

When no external force acts on a system, the total momentum of the system is conserved. The total initial momentum of the system is equal to the total final momentum of the system.

The pitcher and the ball are initially at rest, therefore, the total initial momentum of the system is zero.

Since no external forces act on the system comprising of pitcher and the ball, the total final momentum of the system is also equal to zero.

If the mass of the pitcher is mp and its speed is vp, the mass of the ball is mb and the ball's speed is vb, then the final momentum of the system of pitcher and the ball is given by,

p=m_pv_p+m_bv_b=0

Therefore,

v_p=-\frac{m_b}{m_p} v_p

Substituet 0.15 kg for mb, 50 kg for mp and 35 m/s for vb.

v_p=-\frac{m_b}{m_p} v_p=-\frac{0.15 kg}{50 kg} (35m/s)=-0.105 m/s

The pitcher has a velocity <u> 0.105 m/s</u> opposite to the direction of the velocity of the ball.

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Answer:

The y-component of the electric force on this charge is F_y = -1.144\times 10^{-6}\ N.

Explanation:

<u>Given:</u>

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where, \hat i,\ \hat j are the unit vectors along the positive x and y axes respectively.

The electric field at a point is defined as the electrostatic force experienced per unit positive test charge, placed at that point, such that,

\vec E = \dfrac{\vec F}{q}\\\therefore \vec F = q\vec E\\=(10.4\times 10^{-9})\times (148.0\ \hat i-110.0\ \hat j)\\=(1.539\times 10^{-6}\ \hat i-1.144\times 10^{-6}\ \hat j)\ N.

Thus, the y-component of the electric force on this charge is F_y = -1.144\times 10^{-6}\ N.

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