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ICE Princess25 [194]
2 years ago
5

When a quarterback throws a football, the throwing arm is slowed down to rest during the deceleration phase. Of the angular velo

city of the throwing arm is 1800 deg/sec at the instant of ball release, then how much would the arm deceleration change if the deceleration was reduced from 2 sec to 0.5 sec
Physics
1 answer:
asambeis [7]2 years ago
3 0

Answer:

The deceleration is 4 times higher for 0.5sec than for 2 sec.

Explanation:

If the deceleration happens from 1800deg/sec to zero in 2 sec we have that the magnitude is:

a_{1}=w/t_{1}=(1800deg/sec)/2sec=900 deg/sec^{2}

if now the time to decelerate is 0.5sec:

a_{2}=w/t_{2}=(1800deg/sec)/0.5sec=3600 deg/sec^{2}

The deceleration is now 4 times higher than the previous situation.

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5. Is greater than mg, always

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y-axis:

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N = \frac{m*g + m*ay}{cos \beta }

If ay is positive or zero, N will be greater than mg. If ay is negative, N will be less than mg.

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3 years ago
A ball is launched vertically with an initial speed of y˙0= 50 m/s, and its acceleration is governed by y¨=-g-cDy˙2, where the a
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Answer:

Explanation:

Given

acceleration is given by

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Also acceleration is given by

a=v\frac{\mathrm{d} v}{\mathrm{d} s}

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\Rightarrow Let -g-0.001v^2=t

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t=-g-2.5

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\int_{0}^{s}ds=\int^{-g}_{-g-2.5}\frac{dt}{0.002t}

s=\frac{1}{0.002}lnt|_{-g}^{-g-2.5}

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when air drag is neglected maximum height reached is

h=\frac{v_0^2}{2g}

h=\frac{50^2}{2\times 9.8}

h=127.55\ m

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