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Vitek1552 [10]
3 years ago
8

Please Please Please help on these 3 questions :) 20 POINTS!! - NO LINKS PLEASE

Physics
1 answer:
GenaCL600 [577]3 years ago
3 0

Answer:

Catapult on the ground: Normal, gravity

Catapult (I'm assuming launching marshmallow): Reaction of Force Applied

Marshmallow: Force Applied

Explanation:

This is the forces that act on a stationary object and a launched object. The catapult may also experience a force friction if your teacher is taking a more practical sense.

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It is known that a shark can travel at a speed of 15 m/s.how far can a shark go in 10 seconds?
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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
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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)

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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:

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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}

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