The question is incomplete. The complete question is :
High-speed stroboscopic photographs show that the head of a 200 g golf club is traveling at 60 m/s just before it strikes a 50 g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at 40 m/s. Find the speed of the golf ball just after impact.
Solution :
We know that momentum = mass x velocity
The momentum of the golf club before impact = 0.200 x 60
= 12 kg m/s
The momentum of the ball before impact is zero. So the total momentum before he impact is 12 kg m/s. Therefore, due to the conservation of momentum of the two bodies after the impact is 12 kg m/s.
Now the momentum of the club after the impact is = 0.2 x 40
= 8 kg m/s
Therefore the momentum of the ball is = 12 - 8
= 4 kg m/s
We know momentum of the ball, p = mass x velocity
4 = 0.050 x velocity
∴ Velocity = 
= 80 m/s
Hence the speed of the golf ball after the impact is 80 m/s.
Answer:
its not moving at a constant velocity because it is slowing down
Explanation:
Answer:
Let's see what to do buddy...
Explanation:
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The body start from rest which means :

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In third second means :
t = 2 -----¢ t = 3 -----¢ ∆t = 1
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We have this equation to find the distance.

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And we're done.
Thanks for watching buddy good luck.
♥️♥️♥️♥️♥️
Answer: 0.123 N/C
Explanation: In order to solve this question we have to use the electric Force on a particle produced by an electric field which is given by:
F=q*E
so E=F/q= -2,58* 10^-6/-2.1*10^-5= 0.123 N/C
Waning Gibbous would be the phase?