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
1 answer:
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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Explanation:
given,
t = 133 min = 133 x 60 = 7980 s
angular speed varies from 570 rpm to 1600 rpm
now,
570 rpm =
= 59.69 rad/s
1600 rpm = =
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using equation of rotational motion
ωf = ωi + αt
167.6 = 59.7 + α x 7980
α x 7980 = 107.9
α = 0.0135 rad/s²
Answer:
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