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suter [353]
3 years ago
6

A force in the negative direction of an x-axis is applied for 17ms to a 0.18 kg ball initially moving at 16.0 m/s in the positiv

e direction of the axis. the force varies in magnitude and results in an impulse of magnitude 4.28 ns. what is the velocity of the ball just after the force is applied? indicate the direction of the velocity by its sign.
Physics
1 answer:
Marizza181 [45]3 years ago
5 0
Impulse = Ft=mΔv => Δv = Ft/m = 4.28/0.18 = 23.78 m/s

But,
Δv = v1-v2, where v1 = initial velocity = 16 m/s, v2 = final velocity

Therefore,
v1 - v2 = 23.78 => v2 = v1 - 23.78 => v2 = 16 - 23.78 = -7.78 m/s

The velocity of ball after the force is 7.78 m/s in the direction of the force.

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The magnitude of this vector is 15

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The magnitude of a vector formula is used to calculate the length for a given vector (say v) and is denoted as |v|. So basically, this quantity is the length between the initial point and endpoint of the vector.

Let vector be = a

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Answer:

Therefore, the revolutions that each tire makes is:

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Explanation:

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