A ball is tied to the end of a cable of negligible mass. The ball is spun in a circle with a radius 2.00 m making 0.700 revolutions per second. What is the centripetal acceleration of the ball?

Horizontal circle?

centacc=w^2 * r= (7*2PI)^2 * 2

To find the centripetal acceleration of the ball, we will use the formula:

Ac = (v^2) / r

Where:
Ac is the centripetal acceleration
v is the velocity
r is the radius

Given information:
Radius, r = 2.00 m
Number of revolutions per second, n = 0.700

To find the velocity, we need to convert the number of revolutions per second to angular velocity (ω), using the conversion formula:

ω = 2πn

Let's calculate ω first:

ω = 2π(0.700)
= 4.398 rad/s

Now, we can find the velocity (v) using the formula:

v = ωr

v = (4.398 rad/s)(2.00 m)
= 8.796 m/s

Finally, we can find the centripetal acceleration (Ac) using the formula:

Ac = (v^2) / r

Ac = (8.796 m/s)^2 / 2.00 m
= 77.274 m/s^2

Therefore, the centripetal acceleration of the ball is 77.274 m/s^2.

To find the centripetal acceleration of the ball, we can use the formula a = (v^2)/r, where a is the centripetal acceleration, v is the velocity, and r is the radius.

First, let's find the velocity of the ball. The ball is spinning at a rate of 0.700 revolutions per second. Since each revolution covers the circumference of a circle, the velocity can be calculated by multiplying the circumference by the number of revolutions per second.

The circumference of a circle is given by 2πr, where r is the radius.

In this case, the radius is 2.00 m. So, the circumference is 2π(2.00) = 4π m.

Now, we can calculate the velocity by multiplying the circumference by the number of revolutions per second.

So, the velocity is 4π m/rev * 0.700 rev/s = 2.8π m/s.

Now that we have the velocity, we can calculate the centripetal acceleration using the formula a = (v^2)/r.

Plugging in the values, we get a = (2.8π m/s)^2 / 2.00 m.

Calculating further, we have a = (7.84π^2) m^2/s^2 / 2.00 m.

Simplifying, we get a = 3.92π^2 m/s^2.

Therefore, the centripetal acceleration of the ball is approximately 3.92π^2 m/s^2.

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