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stira [4]
4 years ago
14

A fair ground ride spins its occupants inside a flying saucer-shaped container. if the horizontal circular path the riders follo

w has an 8.00 m radius, at how many revolutions per minute will the riders be subjected to a centripetal acceleration whose magnitude is 1.50 times that due to gravity?
Physics
1 answer:
Licemer1 [7]4 years ago
6 0
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
Below is the solution:

 <span>centripetal accel = 1.5*g 
ω²r = 1.5*9.8m/s² 
ω² * 8m = 14.7 m/s² 
ω = 1.36 rad/s * 1rev/2πrads * 60s/min = 12.9 rpm</span>

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What is the maximum velocity that a 0.3–kilogram mass attached to a 0.75–meter string can have if the mass is whirled around in
Novay_Z [31]
Since this is a horizontal path, we can neglect the force of gravity acting on the body. So in this case we have that the force of tension is equal to the centripetal force, because we have a circular path. 

Fcp=T, where T is the force of tension and Fcp is the centripetal force. 

m*(v²/R)=250 N, where m is the mass of the body and it is m=0.3 kg, v is the max speed of the body, and that is what we are looking for and R is the max length of the string and it is R=0.75 m. 

We divide by m and multiply by R and we get:

v²=(250*R)/m, take the square root:

v=√((250*R)/m)=25 m/s

So the max speed of the body if the max tension is T= 250 N and its max length is R=0.75 m is V=25 m/s. 
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4 years ago
A professional quarterback throws a 0.40 kg football. what is the force of weight?
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Answer:

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what does it mean for a plate to subduct?

Explanation:

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