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antoniya [11.8K]
3 years ago
10

A constant torque is applied to a rigid wheel whose moment of inertia is 2.0 kg · m2 around the axis of rotation. If the wheel s

tarts from rest and attains an angular velocity of 25.0 rad/s in 13.0 s, what is the applied torque (in N · m)? (Indicate the direction with the sign of your answer.)
Physics
1 answer:
Jlenok [28]3 years ago
4 0

Answer:

The applied torque is 3.84 N-m.      

Explanation:

Given that,

Moment of inertia of the wheel is 2\ kg-m^2

Initial speed of the wheel is 0 (at rest)

Final angular speed is 25 rad/s

Time, t = 13 s

The relation between moment of inertia and torque is given by :

\tau=I\alpha \\\\\tau=I\times \dfrac{\omega_f}{t}\\\\\tau=2\times \dfrac{25}{13}\\\\\tau=+3.84\ N-m

So, the applied torque is 3.84 N-m.

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A 2500-kg car is being pushed up a hill at an angle of 35 degrees. Determine the gravitational
velikii [3]

Answer:

The force that pulls the car down is Wₓ = 14052.6 N and the one that pushes the car up is F = 26552.6 N

Explanation:

For this exercise we will use Newton's second law, let's set a reference system where the x axis is parallel to the plane, in the adjoint we can see the forces in the system.

           sin  35 = Wₓ / W

            cos 35 = W_y / W

            Wₓ = W sin 35

            W_y = W cos 35

            Wₓ = 2500 9.8 sin 35

             Wₓ = 14052.6 N

let's write the equations for each axis

and

Y axis  

       N-W_y = 0

       N = W_y

X axis  

       F -Wₓ = m a

       F = Wₓ + m a = mg sin 35 + m a

       F = m (a + g sin 35)

   

let's calculate

       F = 2500 (5 + 9.8 sin 35)

       F = 26552.6 N

The force that pulls the car down is Wₓ = 14052.6 N and the one that pushes the car up is F = 26552.6 N

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3 years ago
Liz rushes down onto a subway platform to find her train already departing. she stops and watches the cars go by. each car is 8.
Snezhnost [94]

 

The average velocity can be calculated using the formula:

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For the 1st car, the velocity is calculated as:

v1 = 8.60 m / 1.80 s = 4.78 m / s

While that of the 2nd car is:

v2 = 8.60 m / 1.66 s = 5.18 m / s

 

Now we can solve for the acceleration using the formula:

v2^2 = v1^2 + 2 a d

Rewriting in terms of a:

a = (v2^2 – v1^2) / 2 d

a = (5.18^2 – 4.78^2) / (2 * 8.6)

a = 0.23 m/s

 

Therefore the train has a constant acceleration of about 0.23 meters per second.

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Explanation:

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