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valentinak56 [21]
3 years ago
7

A pitcher delivers a fast ball with a velocity of 43 m/s to the south. the batter hits the ball and gives it a velocity of 51 m/

s to the north. what was the average acceleration (magnitude and direction) of the ball during the 1.0 ms when it was in contact with the bat?
Physics
1 answer:
hoa [83]3 years ago
4 0
Assuming north as positive direction, the initial and final velocities of the ball are:
v_i=-43 m/s (with negative sign since it is due south)
v_f=+51 m/s
the time taken is t=1.0 ms=0.001 s, so the average acceleration of the ball is given by
a= \frac{v_f-v_i}{t}= \frac{51 m/s-(-43 m/s)}{0.001 s}=9.4 \cdot 10^4 m/s^2
And the positive sign tells us the direction of the acceleration is north.
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From the question we are told that  

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Generally the magnetic field generated by the current in the loop is mathematically represented as

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Now for the earth's magnetic field to be canceled out the magnetic field generated by the loop must be equal to the magnetic field out the earth

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=>     B_e =  \frac{\mu_o  *  N  *  I  }{ 2 * r}

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