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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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A ball is thrown at a 60.0° angle above the horizontal across level ground. It is released from a
yawa3891 [41]
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6 0
3 years ago
The froghopper Philaenus spumarius is supposedly the best jumper in the animal kingdom. To start a jump, this insect can acceler
Kay [80]

Answer:

Part a)

v_f = 4 m/s

Part b)

t = 0.001 s

Part c)

d = 0.815 m

Explanation:

Part a)

As we know that initially the grass hopper is at rest at the ground position

Now the acceleration is given as

a = 4000 m/s^2

distance of the legs that it stretched is given as

s = 2.0 mm

so we have

v_f^2 - v_i^2 = 2 a d

v_f^2 - 0 = 2(4000)(0.002)

v_f = 4 m/s

Part b)

time taken to reach this speed is given as

v_f - v_i = at

4 - 0 = 4000 t

t = 0.001 s

Part c)

as the grass hopper reach the maximum height its final speed would be zero

so we will have

v_f^2 - v_i^2 = 2 a d

0 - 4^2 = 2(-9.81) d

d = 0.815 m

5 0
3 years ago
Protons, neutrons, and electrons are the 3 particles that make up an atom.<br> True or False
bekas [8.4K]
True

Hope that helps
4 0
2 years ago
Convert 543 g to kg (Use correct number of significant figures for answer).
andrew11 [14]

Answer:

0.543 kilogram

Explanation:

thats the answer

5 0
2 years ago
The sampling rate of an ADC is 8.1 kHz. What will be an appropriate cut-off frequency (break frequency) for the anti-aliasing fi
KiRa [710]

Answer:

4000 Hz

Explanation:

An anti-alias filter is usually added in front of the ADC to limit a certain range of input frequencies in order to avoid aliasing. This filter is usually a low pass filter that passes low frequencies but attenuates the high frequencies.

The Nyquist sampling criteria states that the sampling rate should be at least twice the maximum frequency component of the desired signal.

Sampling rate = 2(max input frequency)

From the relation we can find out the cut-off frequency for the anti-aliasing filter.

max input frequency = sampling rate/2

max input frequency = 8100/2 = 4050 Hz

Therefore, 4000 Hz would be an appropriate cut-off frequency for the anti-aliasing filter.

3 0
3 years ago
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