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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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2 years ago
A particle whose speed is 50 m/sec moves along the line from A(2,1) to B (9,25)
WINSTONCH [101]

First, calculate for the distance between the given points A and B by using the equation,

<span>                                                D = sqrt ((x2 – x1)2 + (y2 – y1)2)</span>

 

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<span>                                                D = sqrt((9 – 2)2 + (25 – 1)2)</span>

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I assume the unknown here is the time it would require for the particle to move from point A to B. This can be answered by dividing the calculated distance by the speed given above.

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What is the momentum of a 0.15 kilgram baseball moving at 20 m/s?
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5 0
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A gamma ray burst produces radiation that has a period of 3.6x10-21 s. What wavelength does this radiation have?
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Answer:

The radiation wavelength is  1.08 X 10⁻¹² m

Explanation:

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λ is the wavelength of gamma ray = ?

F is the frequency of the gamma ray = 1/T

T is the period of radiation = 3.6x10⁻²¹ s

\frac{1}{T}  = \frac{c}{\lambda}

λ = T*C

λ = 3.6x10⁻²¹ *  3 x 10⁸

λ =  1.08 X 10⁻¹² m

Therefore, the radiation wavelength is  1.08 X 10⁻¹² m

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The magnitude of the magnetic field at a certain distance from a long, straight conductor is represented by B. What is the magni
masya89 [10]

Answer:

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