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VARVARA [1.3K]
3 years ago
12

Forces of 11.9 N north, 19.1 N east, and 14.4 N south are simultaneously applied to a 3.77 kg mass as it rests on a frictionless

air table. What is the magnitude of its acceleration?
Physics
1 answer:
oksian1 [2.3K]3 years ago
4 0
Force Vector = <19.1i+11.9j-14.4j>N
Force Vector = <19.1i-2.5j>N
Acceleration Vector =(Force Vector)/mass
Acceleration Vector =<19.1i-2.5j>N/3.77
Acceleration Vector =<5.0663i-0.6631j>m/s^2
|a| = (5.0663^2+(-0.6631)^2)^0.5
|a| = 5.11 (m/s^2)
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3 years ago
A child on a swing sweeps out a distance of 45 ft on the first pass. If she is allowed to continue swinging until she​ stops, an
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d = 90 ft

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3 years ago
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Body 1 has a mass m, and its moving in a circle with a radius r at a speed v. It has a centripetal force. F acting on it. If bod
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Answer:

The centripetal force on body 2 is 8 times of the centripetal force in body 1.

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So, the centripetal force on body 2 is 8 times of the centripetal force in body 1.

8 0
3 years ago
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