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Sati [7]
3 years ago
11

Two ice skaters stand together as illustrated in the attached figure. They "push off" and travel directly away from each other,

the boy with a velocity of v = +0.50 m/s. If the boy weighs 710 N and the girl 480 N, what is the girl's velocity in m/s after they push off? (Consider the ice to be frictionless.)
Physics
1 answer:
Tema [17]3 years ago
6 0
By definition, the momentum is given by:
 p = m * v
 Where,
 m = mass
 v = speed.
 On the other hand,
 F = m * a
 Where,
 m = mass
 a = acceleration:
 For the boy we have:
 p1 = m * v
 p1 = (F / a) * v
 p1 = ((710) / (9.81)) * (0.50)
 p1 = 36.19 Kg * (m / s)
 For the girl we have:
 p2 = m * v
 p2 = (F / a) * v
 p2 = ((480) / (9.81)) * (v)
 p2 = 48.93 * v Kg * (m / s)
 Then, we have:
 p1 + p2 = 0
 36.19 + 48.93 * v = 0
 Clearing v:
 v = - (36.19) / (48.93)
 v = -0.74 m / s (negative because the velocity is in the opposite direction of the boy's)
 Answer:
 the girl's velocity in m / s after they push off is -0.74 m / s
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<u>Answer:</u> The number of photons are 3.7\times 10^8

<u>Explanation:</u>

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Putting values in above equation, we get:

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q=165g\times 4.184J/g.^oC\times 77^oC\\\\q=53157.72J

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n=\frac{q}{E}

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