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ICE Princess25 [194]
3 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]3 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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An old manuscript reveals that a landowner in the time of King Arthur held 3.00 acres of plowed land plus a livestock area of 25
iragen [17]

Answer:

(a) Total area is 14.5 roods

(b) Total area is 14674.522 square meters

Explanation:

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Total area = 480 + 100 = 580 square perches

1 rood = 4 perches by 1 perch = 4 square perches

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(b) Total area = 580 square perches

1 perch = 16.5ft = 16.5/3.2808 = 5.03 meters

580 square perches × (5.03 meters/1 perch)^2 = 580 ×25.3009 square meters = 14674.522 square meters

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3 years ago
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