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kotykmax [81]
3 years ago
5

A 0.40 kg ball, attached to the end of a horizontal cord, is rotated in a circle of radius 1.0 m on a frictionless horizontal su

rface. if the cord will break when the tension in it exceeds 64 n, what is the maximum speed the ball can have?
Physics
1 answer:
DerKrebs [107]3 years ago
3 0
<span>The centripetal force for such an arrangement can be found through the equation Fc = mv^2/r where m is the mass of the rotating object, v is that object's velocity, and r is the radius of rotation. In this case, we know that the maximum Fc that can be tolerated by the cord is 64N. Thus we set the equation up and solve for the value of v for which Fc = 64.
   64 = 0.4*(v^2)/1
 64/0.4 = 160 =
v^2 v = sqrt(160) = 12.65 m/s

   At any speed faster than 12.65 m/s, the cord will break.</span>
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PLEASE ANYONE SOLVE THIS NOW FAST PLEASE IM IN A HURRY
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Hello,

<u>Solution for A:</u>

Force = 3.00N

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According to newton's second law of motion;

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<u>Solution for B:</u>

The net force = 4.00N - 3.00N = 1.00N to the left

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Time = 3.00 Seconds

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