A 0.40kg ball, attached to the end of a horizontal cord, is rotated in a circle of radius 2.1m on a frictionless horizontal surface.If the cord will break when the tension in it exceeds 80N , what is the maximum speed the ball can have?

M V^2/R = 80 N

R = 2.1 m
M = 0.40 kg

Solve for V

Well, if the cord breaks when the tension exceeds 80N, we need to figure out the maximum tension in the cord. To do that, we can use the centripetal force formula: F = m * v^2 / r, where F is the tension, m is the mass of the ball, v is the velocity, and r is the radius of the circle.

Now, let's solve for v. Rearranging the formula, we get v = sqrt(F * r / m).

Plugging in the values, we have v = sqrt(80 N * 2.1 m / 0.40 kg). Crunching those numbers, the maximum speed the ball can have is v ≈ 8.18 m/s.

So, watch out for that speed! Too fast and it's clown town for the cord!

To find the maximum speed of the ball, we can equate the tension in the cord to the centripetal force acting on the ball.

The formula for centripetal force is:

Fc = m * v^2 / r

Where:
Fc is the centripetal force
m is the mass of the ball (0.40 kg)
v is the velocity (speed) of the ball
r is the radius of the circle (2.1 m)

Since we know that the tension in the cord cannot exceed 80 N, we can set the centripetal force equal to 80 N:

80 N = (0.40 kg) * v^2 / (2.1 m)

To find the maximum speed, we need to solve this equation for v.

First, we multiply both sides of the equation by (2.1 m):

80 N * (2.1 m) = (0.40 kg) * v^2

Then, we divide both sides of the equation by (0.40 kg):

(80 N * 2.1 m) / 0.40 kg = v^2

Simplifying the equation further:

v^2 = (168 N m) / 0.40 kg

v^2 = 420 N m / kg

Finally, we take the square root of both sides of the equation to find the maximum speed:

v = sqrt(420 N m / kg)

Using a calculator, we can find the maximum speed:

v ≈ 9.165 m/s

Therefore, the maximum speed the ball can have is approximately 9.165 m/s.

To determine the maximum speed of the ball, we need to consider the tension in the cord as it rotates in a circle. The maximum tension the cord can withstand is given as 80 N.

We know that the tension in the cord (T) is responsible for providing the centripetal force (Fc) required to keep the ball in circular motion.

The centripetal force is given by the formula: Fc = (m * v^2) / r

Where:
m = mass of the ball = 0.40 kg
v = velocity of the ball
r = radius of the circle = 2.1 m

We can rearrange the formula to solve for the maximum speed (v):
v^2 = (Fc * r) / m

Substituting the given values:
v^2 = (80 N * 2.1 m) / 0.40 kg

Simplifying the equation:
v^2 = 420 Nm / 0.40 kg
v^2 = 1050

To find the maximum speed, we take the square root of both sides:
v = √1050 m/s

Using a calculator, we can evaluate the square root of 1050, which equals approximately 32.40 m/s.

Therefore, the maximum speed the ball can have is approximately 32.40 m/s.

1.2 m/s