Physics, 02.03.2020 17:17 megababe04

A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.167 kg and moves at v = 4.84 m/s. The circular path has a radius of R = 0.97 m.
What is the minimum velocity so the string will not go slack as the ball moves around the circle?

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Answer from: vickygloom

The minimum velocity of the ball so the string will not go slack as the ball moves around the circle = 6.17 m/s

Explanation:

During the motion of the ball round the circle, the force keeping the ball in circular motion is directed towards the centre of the circle and the gravity is always pointing down.

At the highest position in the motion of the ball, the velocity is at a minimum because the net force responsible for motion is the difference between centripetal force (force in the string, keeping the ball in circular motion) and the weight of the ball.

At the bottom position of the ball, it has maximum velocity during the circular motion (net force = centripetal force + weight).

If the velocity of the ball is too low at the bottom, the ball will not have enough energy to make it to the top.

As the ball comes to the bottom of its circular motion, its kinetic energy increases because its potential energy is decreasing and as the ball moves back to the top of the circle of the circular motion, its kinetic energy decreases because its potential energy is increasing.

So, with respect to the height of the bottom of the circular motion as point 0, the diameter of the circular motion is the maximum vertical height reached by the ball.

And the minimum kinetic energy to still be able to rise to that vertical height = potential energy at that vertical height

(1/2) mv² = mgh

m = 0.167 kg

v = ?

g = 9.8 m/s²

h = 2 × radius = 2 × 0.97 = 1.94 m

(1/2) v² = 9.8 × 1.94

v² = 2×9.8×1.94 = 38.024

v = 6.17 m/s

Answer from: Quest

the unit of the measure = $\frac{j}{kgk}$

explanation:

the property that is quantified by the energy needed per of material to raise the temperature by one.

si unit of energy = joule (j)

si unit of temperature = kelvin (k)

si unit of mass of material = kilogram (kg)

the property that is quantified by the energy needed per of material to raise the temperature by one is called specific heat.

the unit of specific heat is $\frac{j}{kgk}$

Answer from: Quest

During the 1800s. hope this !

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