Math  /  Algebra

Question40. (II) A 0.45kg0.45-\mathrm{kg} ball, attached to the end of a horizontal cord, is revolved in a circle of radius 1.3 m on a frictionless horizontal surface. If the cord will break when the tension in it exceeds 75 N , what is the maximum speed the ball can have?

Studdy Solution
Solve for the maximum speed.
First, isolate v2 v^2 by multiplying both sides by 1.3 1.3 :
75×1.3=0.45×v2 75 \times 1.3 = 0.45 \times v^2
97.5=0.45×v2 97.5 = 0.45 \times v^2
Next, divide both sides by 0.45 0.45 :
v2=97.50.45 v^2 = \frac{97.5}{0.45}
v2=216.67 v^2 = 216.67
Finally, take the square root of both sides to solve for v v :
v=216.67 v = \sqrt{216.67}
v14.72m/s v \approx 14.72 \, \text{m/s}
The maximum speed the ball can have is approximately:
14.72m/s \boxed{14.72 \, \text{m/s}}

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