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Rzqust [24]
3 years ago
5

Moment of inertia Wheel If you lift the front wheel of a poorly maintained bicycle off the ground and then start it spinning at

0.72rev/s, friction in the bearings causes the wheel to stop in just 12s. If the moment of inertia of the wheel about its axle is 0.30kg*m2, what is the magnitude of the frictional torque? Please show your work
Physics
1 answer:
Vinil7 [7]3 years ago
6 0

Answer:

0.113097335529 Nm

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

a = Acceleration

I = Moment of inertia = 0.3 kgm²

\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{0-0.72\times 2\pi}{12}\\\Rightarrow \alpha=-0.376991118431\ rad/s^2

Torque is given by

\tau=I\alpha\\\Rightarrow \tau=0.3\times (-0.376991118431)\\\Rightarrow \tau=-0.113097335529\ Nm

The magnitude of torque is 0.113097335529 Nm

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Answer:

Weight is what you get when a certain amount of gravity is acting on that mass, and something, like the surface of a planet, is resisting that action. In space, when falling freely, there's nothing resisting the pull of gravity so weight disappears. Mass however stays.

hope this helps u

Explanation:

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A projectile is launched into the air with the initial speed of vi = 40 m/s at a launch angle of 20 degrees above the horizontal
Sphinxa [80]

The range of the projectile is 188 m

Explanation:

The motion of the arrow in this problem is a projectile motion, so it follows a parabolic path which consists of two independent motions:  

- A uniform motion (constant velocity) along the horizontal direction  

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction  

The path of a projectile is the combination of these two motions: see figure in attachment.

In order to find the horizontal range of the projectile, we just need to calculate the horizontal distance travelled.

We have:

t = 5.0 s (time of fligth of the projectile)

and the horizontal velocity is constant, and it is given by

v_x = v_i cos \theta

where

v_i = 40 m/s is the initial velocity

\theta=20^{\circ} is the angle of projection

Substituting,

v_x = (40)(cos 20^{\circ})=37.6 m/s

And therefore, the range of the projectile is:

d=v_x t = (37.6)(5.0)=188 m

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

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