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Masja [62]
3 years ago
14

A 0.7 kg block attached to a spring with force constant 160 Nm is free to move on a frictionless, horizontal surface. The block

is released from rest when the
spring is stretched + 0.15 m as shown in figure below. At the instant the block is released the force vector on the block is.
(Consider Right direction is positive direction and Left direction is negative direction)
- 34.29 N
+ 34.29 N
+ 24 N
Оe
h Oa-241
Physics
1 answer:
Tom [10]3 years ago
3 0

Answer:

+ 24 N

Explanation:

the computation is shown below:

Given that

Mass of the block = m = 0.7 kg

Sprint constant = k = 160 N / m

x = 0.15m

Now the force on the block is

F = kx

= (160) (0.15)

= 24 N

As the instant block is released so the acting of the force on the block is positive and it would be in a positive direction i.e. right direction

Therefore the third option is correct

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Answer:

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Jerome solves a problem using the law of conservation of momentum. What should Jerome always keep constant for each object after
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Jerome solves a problem using the law of conservation of momentum. What should Jerome always keep constant for each object after the objects collide and bounce apart?

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Answer:

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Explanation:

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Two male moose charge at each other with the same speed and meet on a icy patch of tundra. as they collide, their antlers lock t
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3 0
4 years ago
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