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4 years ago

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A girl moves quickly to the center of a spinning merry-go-round, traveling along the radius of the merry-go-round. Which of the

following statements are true?Check all that apply.Check all that apply.The angular speed of the system increases.The moment of inertia of the system remains constant.The angular speed of the system decreases.The moment of inertia of the system increases.The moment of inertia of the system decreases.The angular speed of the system remains constant.

1 answer:

DochEvi [55]4 years ago

3 0

Answer:

The angular speed of the system increases.

The moment of inertia of the system decreases.

Explanation:

As we know that the girl is going towards the center of the circle so here the moment of inertia of the girl is given as

$I = mr^2$

here we know that

r = position of the girl from the center of the disc

now we know that the girl is moving towards the center so its distance will continuously decreasing

So the moment of inertia of the girl will decrease

Now we know that that with respect to the center of the disc there is no torque on the disc + girl system

So here we can use angular momentum conservation

So we have

$I\omega = constant$

since moment of inertia is decreasing for the system

so angular speed will increase

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amid [387]

Answer:

Part a)

$t = 1.65 s$

Part b)

$x = 40.4 m$

Since the distance of other building is 15 m so YES it can make it to other building

Part c)

$v = 27.3 m/s$

direction of velocity is given as

$[tex]\theta = 26.35 degree$

Explanation:

Part a)

acceleration due to gravity on this planet is 3/4 times the gravity on earth

So the acceleration due to gravity on this new planet is given as

$a = \frac{3}{4}(9.81)$

$a = 7.36 m/s^2$

now the vertical displacement covered by the canister is given as

$y = 10 m$

now by kinematics we have

$y = \frac{1}{2}gt^2$

$10 = \frac{1}{2}(7.36)t^2$

$t = 1.65 s$

Part b)

Horizontal speed of the canister is given as

$v_x = 24.5 m/s$

now the distance moved by it

$x = v_x t$

$x = 24.5 (1.65)$

$x = 40.4 m$

Since the distance of other building is 15 m so YES it can make it to other building

Part c)

Final velocity in X direction will remains the same

$v_x = 24.5 m/s$

final velocity in Y direction

$v_y = v_i + at$

$v_y = 0 + (7.36)(1.65)$

$v_y = 12.14 m/s$

now magnitude of velocity is given as

$v = \sqrt{v_x^2 + v_y^2}$

$v = \sqrt{24.5^2 + 12.14^2}$

$v = 27.3 m/s$

direction of velocity is given as

$\theta = tan^{-1}\frac{v_y}{v_x}$

$\theta = tan^{-1}\frac{12.14}{24.5}$

$[tex]\theta = 26.35 degree$

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Answer:

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A fire hose held near the ground shoots water at a speed of 6.5 m/s. At what angle(s) should the nozzle point in order that the

valina [46]

The speed of water can be split into vertical and horizontal speed components:
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MAXImum [283]

Answer:

d. The magnitude of the work done by the earth on the satellite is non zero

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