Which hand is negatively charged? ОА ОВ. Ос. OD

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

Over [174]

What Kepler's constant ? ? ! ?

The only constant in Kepler's laws is in the third one, where it says something to the
effect that (square of a body's period) / (cube of its distance from the central body)
is a constant.

That means it's a constant for multiple little ones orbiting the same central body.
But it's not the same constant for other central bodies.

It's one constant for the planets, asteroids, and comets orbiting the sun.

It's a different constant for the moon, TV satellites, weather satellites,
and military satellites orbiting the Earth.

4 0

3 years ago

15 points. give me the method.

AveGali [126]

Answer:

$\boxed{{160 \: m(s)}^{ - 1} }$

Explanation:

$if \:the \: frequencies \: are \to \\ f_{1} = 640Hz \\ and \\f_{2} = 480Hz \: \\ but \: \boxed{v = f \gamma }: f = \frac{v}{ \gamma } \\ if \: \gamma_{1} - \gamma _{2} = 1 = \gamma \\ f_{1} - f_{2} = 640 - 480 = \boxed{ 160Hz} = f \\ v = f \gamma = 160 \times 1 = \boxed{{160 \: m(s)}^{ - 1} }$

5 0

2 years ago

You are riding in the passenger seat of a car as it goes around a tight turn. You slide across the seat to the passenger side do

DochEvi [55]

Answer:

1. b. The door is exerting a centripetal force on you that balances the centrifugal force of the turn.

2. b. There is no net force acting on the object.

Explanation:

1. This is because as you move to the right due to the centrifugal force of the turn, a corresponding centripetal force acts on you due to the door which does not allow you fall out of the car since, the door is exerting a centripetal force on you that balances the centrifugal force of the turn.

So, the answer is b

2. This is because, since the object moves at a constant speed and thus does not accelerate, no net force can act on it since, a net force would imply that the object accelerates. Note that a constant speed does not imply that no force acts on it. It only shows that the resultant or net force is zero since the object does not accelerate.

So, there is no net force acting on the object.

So, b is the answer.

6 0

2 years ago

Which of the following statements are examples of positive peer pressure?

Crazy boy [7]

Answer:

C and D

Explanation:

have a great day

5 0

2 years ago

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