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Delvig [45]
3 years ago
8

In short-track speed skating, the track has straight sections and semicircles 16 m in diameter. assume that a 64 kg skater goes

around the turn at a constant 13 m/s. what is the horizontal force on the skater?
Physics
1 answer:
Alenkasestr [34]3 years ago
3 0
The diameter of the semicircular portion of the track is 16 m.
Therefore its radius is
r = 8 m.

The tangential velocity of the skater is 13 m/s.
Therefore the angular speed is
ω = v/r = (13 m/s)/(8 m) = 1.625 rad/s

The horizontal force on the skater is due to centripetal acceleration of
a = r*ω² = (8 m)*(1.625 rad/s)² = 21.125 m/s².

The force acting on the 64-kg skater is
F = m*a = (64 kg)*(21.125 m/s²) = 1352 N

Answer: 1352 N

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(ma×va) + (mb×vb) = (ma×va') + (mb×vb')

(1,783×va) + (1,600×0) = (1,783×8) + (1,600×8)

(1,783×va) + 0 = 14,264+12,800

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va \:=\: \frac{27,064}{1.783}

va = 15.18 m/s

Learn more about The law of momentum conservation here: brainly.com/question/7538238

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