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taurus [48]
3 years ago
14

A wheel with rotational inertia i is mounted on a fixed, frictionless axle. the angular speed ω of the wheel is increased from z

ero to ωf in a time interval t. what is the average net torque τ on the wheel during this time interval?
Physics
1 answer:
Annette [7]3 years ago
4 0
According to Newton's second law:
τ = Iα, where I is the moment of inertia and α is the angular acceleration.
Angular acceleration can be written as:
α = (ωf - ωi) / t
α = ωf/t

Substituting into the second law:
τ = Iωf/t
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Your friend is constructing a balancing display for an art project. She has one rock on the left ( ms=2.25 kgms=2.25 kg ) and th
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Torque = 35.60 N.m (rounded off to 3 significant figures.

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5 0
3 years ago
A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 38 ft/s2. what is the dist
e-lub [12.9K]

Convert 38 ft/s^2 to mi/h^2. Then we se the conversion factor > 1 mile = 5280 feet and 1 hour = 3600 seconds.

So now we show it > 38  \frac{ft}{s^2}  x  \frac{1mi}{5280ft} x  \frac{(3600s)^2}{(1h)^2} = 93272.27  \frac{mi}{h^2}

Then we have to use the formula of constant acceleration to determine the distance traveled by the car before it ended up stopping.

Which the formula for constant acceleration would be > v_2^2=v_1^2 + 2as

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When it stops the final velocity is (v_2=0)

Since the given is deceleration it means the number we had gotten earlier would be a negative so a = -93272.27

Then we substitute the values in....

0^2 = 50^2 + 2(-93272.27)s&#10;&#10;0 = 2500 - 186544.54s&#10;&#10;Isolate S next.&#10;&#10;185644.54s= 2500&#10;&#10;s =  2500/(185644.54)&#10;&#10;s=0.0134&#10;

So we can say the car stopped at 0.0134 miles before it came to a stop but to express the distance traveled in feet we need to use the conversion factor of 1 mile = 5280 feet in otherwards > 0.0134 mi *  \frac{5280ft}{1mi}  = 70.8 ft
So this means that the car traveled in feet 70.8 ft before it came to a stop.

4 0
3 years ago
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