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salantis [7]
2 years ago
12

if a 0.040-kg stone is whirled horizontally on the end of a 60-m string at a speed of 4.4 m/s, what is the centripetal force ? *

0.013 n 4 n 0. 32 n 20.1 n
Physics
1 answer:
SSSSS [86.1K]2 years ago
4 0

Answer:

0.013N

Explanation:

F=\frac{mv^{2} }{r}

m=0.04, v=4.4m/s, r=60

F=0.013 N

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Hope this helps!

3 0
3 years ago
A child bounces a 48 g superball on the sidewalk. the velocity change of the superball is from 26 m/s downward to 17 m/s upward.
Nataly_w [17]
By definition we have the momentum is:
 P = m * v
 Where,
 m = mass
 v = speed
 Before the impact:
 P1 = (0.048) * (26) = 1.248 kg * m / s
 After the impact:
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 Then we have that deltaP is:
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 deltaP = (- 0.816) - (1,248)
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 Clearing F:
 F = (deltaP) / (delta t)
 Substituting the values
 F = (- 2.064) / (1/800) = - 1651.2N
 answer:
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