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Lesechka [4]
3 years ago
5

Two ice skaters stand together. They push off and travel directly away from each other, the boy with a velocity of v = +0.35 m/s

. If the boy weighs 745N and the girl 477 N, what is the girl's velocity after they push off? ( Consider the ice to be frictionless)_____m/s
Physics
1 answer:
DIA [1.3K]3 years ago
5 0

Answer:

-0.55m/s

Explanation:

Given that: For the boy

Weight = 745N

Velocity = +0.35 m/s

Mass of the boy = ?

g = 9.81m/s^2

W = mg

745 = m×9.81

m = 75.94kg

For the girl

Given that:

Weight = 477 N

g = 9.81m/s^2

m = ?

W = mg

477 = m×9.81m/s^2

m = 48.62kg

To solve for the v of the girl, the two has to add up

48.62kg×v + 75.94kg×+0.35 m/s = 0

48.62v + 26.579 = 0

48.62v = - 26.579

v = -26.579/48.62

v = -0.5466

v = -0.55m/s

Hence, the velocity of the girl is -0.55m/s.

The negative sign is as a result of the two of them moving is opposite direction.

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