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3 years ago

Draw a net force arrow on the picture below. What is the net force? State the direction.

Physics

2 answers:

zalisa [80]3 years ago

6 0

Answer:

Net force=0 N

Explanation:

The drawing will look like the attached image, with no net force on the object. We stablished a reference system located in the center of the body, with + axis take right and upward for x and y. All forces are on the x axis, there is one positive force of 7 N (pointing right) and to negative forces of -4 and -3 N (pointing left) if we add the vectors in the x axis we have that

+7-4-3=0 N

The net force is then 0 newton and the drawing has no net force vector then

exis [7]3 years ago

5 0

Answer: No arrow; no net force.

Explanation: -7 + 7= 0 same thing here

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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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2 years ago

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Rama09 [41]

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3 years ago

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gtnhenbr [62]

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Explanation:

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2 years ago

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LekaFEV [45]

Answer and Explanation:

The butcher will balance the meat and the 0.5 kg mass on one side of the beam balance and 2 kg mass on the other side of the beam balance.
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2 years ago

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stira [4]

Answer:

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Applying

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