For every action force, there is: A. a net force. B. an unbalanced force. C. a friction force. D. an equal and opposite force. E
1 answer:
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Explanation:
It is given that,
Mass of golf club, m₁ = 210 g = 0.21 kg
Initial velocity of golf club, u₁ = 56 m/s
Mass of another golf ball which is at rest, m₂ = 46 g = 0.046 kg
After the collision, the club head travels (in the same direction) at 42 m/s. We need to find the speed of the golf ball just after impact. Let it is v.
Initial momentum of golf ball,
After the collision, final momentum
Using the conservation of momentum as :
v = 63.91 m/s
So, the speed of the golf ball just after impact is 63.91 m/s. Hence, this is the required solution.
Answer:
Explanation:
Let the extension in the spring be x .
restoring force = weight of block
kx = mg
x =
= 23.84 cm
b )
When the elevator is going upwards
Restoring force = mg + ma
k x₁ = 10.9 ( 9.8 + 1.89 )
x₁ = 28.44 cm
( y coordinate will be - ( 28.44 - 23.84 ) = - 4.6 cm )
c ) When the cable snaps , both elevator and block undergo free fall . In this case apparent g = 0
Since the spring is stretched by 28.44 cm , a restoring force continues to act on the block which is equal to
.2844 x 448
= 127.41 N
So a net acceleration a will act on the block
a = 127.41 / 10.9
= 11.68 m / s²
The block will undergo SHM with amplitude equal to 28.44 cm .