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WINSTONCH [101]

2 years ago

5

forces F1 (east) and F2 are simultaneously applied to a 3.0kg mass. when F2 is east, a=5.0m/s² east and when F2 is west a=1.0m/s

² east. Determine the magnitude of F1 and F2.

1 answer:

zhenek [66]2 years ago

5 0

Answer:

345453-5676

its the right answer

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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

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:

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

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What is the magnitude momentum of a 2 kg body that moves with a speed of 3 m/s? (1 point)

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Answer:

<h2>D</h2>

Explanation:

The momentum of an object can be found by using the formula

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From the question

mass = 2 kg

velocity/speed = 3 m/s

We have

momentum = 2 × 3 = 6

We have the final answer as

<h3>6 kg.m/s</h3>

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Answer:

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