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ziro4ka [17]
3 years ago
11

A 240 g toy car is placed on a narrow 60-cm-diameter track with wheel grooves that keep the car going in a circle. The 1.0 kg tr

ack is free to turn on a frictionless, vertical axis. The spokes have negligible mass. After the car's switch is turned on, it soon reaches a steady speed of 0.79 m/s relative to the track. What then is the track's angular velocity, in rpm?
Physics
1 answer:
lesya [120]3 years ago
8 0

Answer:

The track's angular velocity is W2 = 4.15 in rpm

Explanation:

Momentum angular can be find

I = m*r^2

P = I*W

So to use the conservation

P1 + P2 = 0

I1*W1 + I2*W2 = 0

Solve to w2 to find the angular velocity

0.240kg*0.30m^2*0.79m/s=-1kg*0.30m^2*W2

W2 = 0.435 rad/s

W2 = 4.15 rpm

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Which of the following is the correct SI unit to measure of rope?
sergij07 [2.7K]

Hello!

The answer should be the third option.

Explanation: Because the meters is the correct SI unit to measure of rope. And thank you for posting at here on Brainly. And have a great day! -Charlie

7 0
3 years ago
What is the weight of a box with a mass of 150 kg on Earth?
Dmitry [639]

Answer:

W = M g = 150 kg * 9.81 m/s^2 = 1470 N

You were only given 3 significant figures in the question.

6 0
2 years ago
Initially, a particle is moving at 5.25 m/s at an angle of 35.5° above the horizontal. Three seconds later, its velocity is 6.0
ivolga24 [154]

Answer:

 a =( -0.32 i ^ - 2,697 j ^)  m/s²

Explanation:

This problem is an exercise of movement in two dimensions, the best way to solve it is to decompose the terms and work each axis independently.

Break down the speeds in two moments

initial

  v₀ₓ = v₀ cos θ

  v₀ₓ = 5.25 cos 35.5

v₀ₓ = 4.27 m / s

   v_{oy} = v₀ sin θ

 v_{oy}= 5.25 sin35.5

v_{oy} = 3.05 m / s

Final

vₓ = 6.03 cos (-56.7)

vₓ = 3.31 m / s

v_{y} = v₀ sin θ

v_{y} = 6.03 sin (-56.7)

v_{y} = -5.04 m / s

Having the speeds and the time, we can use the definition of average acceleration that is the change of speed in the time order

    a = (v_{f} - v₀) /t

    aₓ = (3.31 -4.27)/3

    aₓ = -0.32 m/s²

    a_{y} = (-5.04-3.05)/3

   a_{y} =  -2.697 m/s²

6 0
3 years ago
Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.20×10^6 N, one at an angle 14.0∘ west of north,
laila [671]

Answer:

1.45544 J

Explanation:

See attachment

5 0
3 years ago
The static frictional force between a 95-kilogram object and the floor is 45 Newtons. The kinetic frictional force is only 22 Ne
Lisa [10]

Answer:

F = 69.5 [N]

Explanation:

We must remember that the friction force is defined as the product of the normal force by the coefficient of friction, and it can be calculated by the following expression.

f=N*miu

where:

N = normal force [N]

miu = friction coefficient

f = friction force = 22 [N]

Now we must calculate the force exerted by means of Newton's second law which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

F - f = m*a

where:

F = force exerted [N]

f = friction force [N]

m = mass = 95 [kg]

a = acceleration = 0.5 [m/s²]

Now replacing:

F - 22 = 95*0.5\\F = 47.5 + 22\\F = 69.5 [N]

6 0
3 years ago
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