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Nitella [24]
3 years ago
13

Doug is driving a golf ball off the tee. His downswing takes 0.50sec from the top of the swing until ball impact. At the top of

the swing, the club's angular velocity is zero; at the instant of ball impact, the club's angular velocity is 30rad/s. The swing moment of inertia of the club about the grip is 0.220kg*m^2. What average torque does Doug exert on the golf club during the downswing?
A. 13.2
B. 26.4
C. 64.6
D. 106.8
Physics
1 answer:
Dmitry [639]3 years ago
6 0
D.eeeeeeezzzzzzzz nuuuuutttttssss!
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Answer:

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Explanation:

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A 0.0450 kg bullet is accelerated from rest to a speed of 425 m/s in a 2.25 kg rifle (which is inititally at rest). The pain of
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Answer:

If the rifle is held loosely away from the shoulder, the recoil velocity will be of -8.5 m/s, and the kinetic energy the rifle gains will be 81.28 J.

Explanation:

By momentum conservation, <em>and given the bullit and the recoil are in a straight line</em>, the momentum analysis will be <em>unidimentional</em>. As the initial momentum is equal to zero (the masses are at rest), we have that the final momentum equals zero, so

0=P_{f}=m_{b} *v_{b}+m_{r}*v_{r}

now we clear v_{r} and use the given data to get that

v_{r}=-8.5\frac{m}{s}

<em>But we have to keep in mind that the bullit accelerate from rest to a speed of 425 m/s</em>, then <u>if the rifle were against the shoulder, the recoil velocity would be a fraction of the result obtained</u>, but, as the gun is a few centimeters away from the shoulder, it is assumed that the bullit get to its final velocity, so the kick of the gun, gets to its final velocity \bold{v_{r}} too.

Finally, using v_{r} we calculate the kinetic energy as

K=\frac{1}{2}m_{r}v_{r}^{2}=81.28J

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