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Vadim26 [7]
3 years ago
10

What is the period? Blank seconds.

Physics
1 answer:
FromTheMoon [43]3 years ago
8 0

Answer:

Period refers to the time for something to happen and is measured in seconds/cycle

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Read 2 more answers
A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the
Margaret [11]

Hi there!

\boxed{\omega = 0.38 rad/sec}

We can use the conservation of angular momentum to solve.

\large\boxed{L_i = L_f}

Recall the equation for angular momentum:

L = I\omega

We can begin by writing out the scenario as a conservation of angular momentum:

I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)

I_m = moment of inertia of the merry-go-round (kgm²)

\omega_m = angular velocity of merry go round (rad/sec)

\omega_f = final angular velocity of COMBINED objects (rad/sec)

I_b = moment of inertia of boy (kgm²)

\omega_b= angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

I = MR^2

We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

\omega = \frac{v}{r}

L_b = MR^2(\frac{v}{R}) = MRv

Plug in the given values:

L_b = (20)(3)(5) = 300 kgm^2/s

Now, we must solve for the boy's moment of inertia:

I = MR^2\\I = 20(3^2) = 180 kgm^2

Use the above equation for conservation of momentum:

600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}

8 0
2 years ago
Please need help with this
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<u>Note that</u>:

The gravitational potential energy = mgh

where m: is the mass, g: the acceleration due to the gravity and h is the height from the earth surface

Then, we can increase the gravitational potential energy by increasing the mass or the height from the earth surface

<u>In our question</u>, we can increase the gravitational potential energy by

<u>A) Strap a boulder to the car so that it wights more.</u>

7 0
3 years ago
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