Answer the question plz

A brave child decides to grab onto an already spinning merry‑go‑round. The child is initially at rest and has a mass of 25.5 kg.

Svetllana [295]

Answer:

72.75 kg m^2

Explanation:

initial angular velocity, ω = 35 rpm

final angular velocity, ω' = 19 rpm

mass of child, m = 15.5 kg

distance from the centre, d = 1.55 m

Let the moment of inertia of the merry go round is I.

Use the concept of conservation of angular momentum

I ω = I' ω'

where I' be the moment of inertia of merry go round and child

I x 35 = ( I + md^2) ω'

I x 35 = ( I + 25.5 x 1.55 x 1.55) x 19

35 I = 19 I + 1164

16 I = 1164

I = 72.75 kg m^2

Thus, the moment of inertia of the merry go round is 72.75 kg m^2.

5 0

3 years ago

a 0.05-kg starts from rest at a height of 0.95m. assuming no friction, what is the kinetic energy of the car when it reaches the

Alex73 [517]

Fhubugukhkbjhvlfipy0iy ft

3 0

3 years ago

What is the distance traversed by the particle between 0 seconds and 6 seconds?

ad-work [718]

Answer:

I dont know to be honest

Explanation:

i dont know to be honest

5 0

3 years ago

The terminal velocity is not dependent on which one of the following properties? the drag coefficient 1 the force of gravity 2 c

ahrayia [7]

<h2>Answer: the falling time</h2>

Explanation:

When a body or object falls, basically two forces act on it:

1. The force of air friction, also called "drag force" $D$ :

$D={C}_{d}\frac{\rho V^{2} }{2}A$ (1)

Where:

$C_ {d}$ is the drag coefficient

$\rho$ is the density of the fluid (air for example)

$V$ is the velocity

$A$ is the transversal area of the object

So, this force is proportional to the transversal area of the falling element and to the square of the velocity.

2. Its weight due to the gravity force $W$ :

$W=m.g$

(2)

Where:

$m$ is the mass of the object

$g$ is the acceleration due gravity

So, at the moment when the drag force equals the gravity force, the object will have its terminal velocity:

$D=W$ (3)

${C}_{d}\frac{\rho V^{2} }{2}A=m.g$ (4)

$V=\sqrt{\frac{2m.g}{\rho A{C}_{d}}}$ (5) This is the terminal velocity

As we can see, there is no "falling time" in this equation.

Therefore, the terminal velocity is not dependent on the falling time.

6 0

3 years ago

How does the motion of atoms relate to hot air balloons?

Neporo4naja [7]

Magic magic and more magic

6 0

3 years ago