A 28-kg beginning roller skater is standing in front of a wall. By pushing against the wall, she is propelled backward with a velocity of -1.2 m/s. While pushing, her hands are in contact with the wall for 0.80 s. Ignoring frictional effects, find the magnitude and direction of the impulse acting on the skater.

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asuwafohjames asuwafohjames · Answer 1 · 2020-01-23T10:12:25+01:00

Answer:

33.6 Ns backward.

Explanation:

Impulse: This can be defined as the product of force and time. The S.I unit of impulse is Ns.

From Newton's second law of motion,

Impulse = change in momentum

I = mΔv................................. Equation 1

Where I = impulse, m = mass of the skater, Δv = change in velocity = final velocity - initial velocity.

Given: m = 28 kg, t = 0.8 s, Δv = -1.2-0 = -1.2 m/s (Note: the initial velocity of the skater = 0 m/s)

Substituting into equation 1

I = 28(-1.2)

I = -33.6 Ns

Thus the impulse = 33.6 Ns backward.

priyankatutor priyankatutor · Answer 2 · 2022-01-12T07:46:40+01:00

Impulse can be defined as the change in momentum. The impulse is 33.6 Ns backward as the sign on impulse is negative.

It can be defined as the product of force and time or a change in momentum. The S.I unit of impulse is Ns.

[tex]I = m\Delta v[/tex]

Where,

[tex]I[/tex]= impulse,

[tex]m[/tex] = mass of the skater = 28 kg

[tex]\Delta v[/tex] = change in velocity = [tex]-1.2-0 =\bold { -1.2 \ m/s }[/tex]

Put the values in the equation,

[tex]I = 28 (-1.2)\\\\I = -33.6 \rm \ Ns[/tex]

Therefore, the impulse is 33.6 Ns backward as the sign on impulse is negative.

To know more about impulse,

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