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2 years ago

A 2kg ball rotates on the end of a 1.4m long string. The ball makes 5 revolutions in 4.4s. What is the speed of the ball?

Physics

1 answer:

nlexa [21]2 years ago

3 0

The speed or the linear velocity of the stone is 9.93 m/s; option C

<h3>What is the speed or linear velocity of the stone?</h3>

The linear velocity of the stone is given by the formula below:

v = wr

where w is angular velocity and r is radius.

w = θ/t

1 complete revolution = 2π radians

5 revolutions = 10π radians

w = 10π/4.4 rad/s

v = 10π/4.4 rad/s * 1.4

v = 9.93 m/s

In conclusion, the linear velocity is obtained from the angular velocity and radius of the string.

Learn more about linear and angular velocity at: brainly.com/question/15154527

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Explanation:

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3 years ago

Sam is playing football. She kicks the ball with an average force of 75 N.

damaskus [11]

Answer:

22.5J

Explanation:

Here the force is given. Also, the displacement is given as 30cm.

First we should check if all the values are in their standard form.

Here 30cm should be converted to metre by dividing it with 100.

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3 years ago

(I) In a ballistic pendulum experiment, projectile 1 results in a maximum height h of the pendulum equal to 2.6 cm. A second pro

Kipish [7]

Answer:

The second projectile was 1.41 times faster than the first.

Explanation:

In the ballistic pendulum experiment, the speed (v) of the projectile is given by:

$v = \frac{m + M}{m} \cdot \sqrt{2gh}$

<em>where m: is the mass of the projectile, M: is the mass of the pendulum, g: is the gravitational constant and h: is the maximum height of the pendulum. </em>

To know how many times faster was the second projectile than the first, we need to take the ratio for the velocities for the projectiles 2 and 1:

$\frac{v_{2}}{v_{1}} = \frac{\frac{m_{2} + M}{m_{2}} \cdot \sqrt{2gh_{2}}}{\frac{m_{1} + M}{m_{1}} \cdot \sqrt{2gh_{1}}}$ (1)

<em>where m₁ and m₂ are the masses of the projectiles 1 and 2, respectively, and h₁ and h₂ are the maximum height reached by the pendulum by the projectiles 1 and 2, respectively. </em>

Since the projectile 1 has the same mass that the projectile 2, we can simplify equation (1):

$\frac{v_{2}}{v_{1}} = \frac{\sqrt{h_{2}}}{\sqrt{h_{1}}}$

$\frac{v_{2}}{v_{1}} = \frac{\sqrt{5.2 cm}}{\sqrt{2.6 cm}}$

$\frac{v_{2}}{v_{1}} = 1.41$

Therefore, the second projectile was 1.41 times faster than the first.

I hope it helps you!

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