A 0.40 kg ball, attached to the end of a horizontal cord, is rotated in a circle of radius 1.5 m on a frictionless horizontal surface. The cord will break when the tension in it exceeds 61 N.

What is the maximum speed the ball can have?

max tension= m*v^2/r

solve for v.

15.12

2341234

61=0.40v^2/1.5

91.5=0.40v^2
228.75=v^2
15.12=v

To find the maximum speed of the ball, we need to consider the tension in the cord at maximum speed. The tension in the cord can be found using the centripetal force formula:

Tension = mass * centripetal acceleration

The centripetal acceleration is given by the formula:

Centripetal acceleration = (velocity^2) / radius

Since we are looking for the maximum speed, we can assume that the tension in the cord is equal to the breaking point (61 N). Therefore, we can rearrange the formulas to solve for velocity:

61 N = mass * [(velocity^2) / radius]

Plugging in the given values:

61 N = 0.40 kg * [(velocity^2) / 1.5 m]

Now, let's solve for the velocity:

61 N * (1.5 m / 0.40 kg) = velocity^2

Simplifying the equation:

220.5 N * kg/m = velocity^2

Now, we can take the square root of both sides to find the maximum velocity:

velocity = sqrt(220.5 N * kg/m)

Calculating the square root:

velocity ≈ 14.85 m/s

Therefore, the maximum speed the ball can have is approximately 14.85 m/s.