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kogti [31]
3 years ago
7

A 4.00 kg ball is swung in a circle on the edge of a 1.50 m rope. The time it takes for the ball to complete one rotation is 3.4

0 s. Please show all work and equation.
a) What is the velocity of the ball?

b) What is the acceleration of the ball?

c) What is the force on the ball?
Physics
1 answer:
lorasvet [3.4K]3 years ago
8 0

Answer:

The answer is below

Explanation:

The length of the rope is equal to the radius of the circle formed by the complete rotation of the rope. Therefore the radius = 1.50 m.

a) The distance covered by the rope when completing one rotation is the same as the perimeter of the circle. Hence:

Distance covered in one rotation = 2π * radius = 2π * 1.5 = 3π meters

The velocity of the ball = Distance / time = 3π meters / 3.4 seconds = 2.77  m/s

b) The initial velocity (u) is 0 m/s, the final velocity is 2.77 m/s during time (t) = 3.4 s. Hence acceleration (a):

v = u + at

2.77 = 3.4a

a = 0.82 m/s²

c) Force on ball = mass * acceleration = 4 * 0.82 = 3.28 N

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A ball rolls down the hill which has a vertical height of 15 m. Ignoring friction what would be the gravitational potential ener
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a) Potential energy: 147 m [J]

The gravitational potential energy of an object is given by

U=mgh

where

m is its mass

g=9.8 m/s^2 is the acceleration of gravity

h is the height of the object above the ground

In this problem,

h = 15 m

We call 'm' the mass of the ball, since we don't know it

So, the potential energy of the ball at the top of the hill is

U=(m)(9.8)(15)=147 m (J)

b) Velocity of the ball at the bottom of the hill: 17.1 m/s

According to the law of conservation of energy, in absence of friction all the potential energy of the ball is converted into kinetic energy as the ball reaches the bottom of the hill. Therefore we can write:

U=K=\frac{1}{2}mv^2

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We know from part a) that

U = 147 m

Substituting into the equation above,

147 m = \frac{1}{2}mv^2

And re-arranging for v, we find the velocity:

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brainly.com/question/28609499

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