A ball on the end of a string is whirled around in a horizontal circle of radius 0.300 m. The plane of the circle is 1.00 m above the ground. The string breaks and the ball lands 2.30 m (horizontally) away from the point on the ground directly beneath the ball's location when the string breaks. Find the centripetal acceleration in m/s2 of the ball during its circular motion.

86.7 m/s²

Explanation:

Given that

Radius of the circle, r = 0.3 m

Height of the plane above the ground, = 1 m

Distance at which the ball lands, d = 2.3 m

To solve this question, we would use one of the principles of projectile motion.

v = ut + 1/2at²

-1 = 0 + 1/2 * -9.8 * t²

-1 = -4.9 t²

t² = 1/4.9

t² = 0.204

t = √0.204

t = 0.45 s

Then again, it's constant horizontal speed is

v(x) = x/t

v(x) = 2.3 / 0.45

v(x) = 5.1 m/s

And finally, the centripetal acceleration is gotten using the formula

a = v²/r

a = 5.1² / 0.3

a = 26.01 / 0.3

a = 86.7 m/s²

answer: i don't know the answer good luck

explanation:

what experimental evidence led scientists to change from the previous model to this one?

a few of the positive particles aimed at a gold foil seemed to bounce back.

the colors of light emitted from heated atoms had very specific energies.

experiments with water vapor showed that elements combine in specific proportions.

cathode rays were bent in the same way whenever a magnet was brought near them.