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3 years ago

9

Consider a bicycle wheel to be a ring of radius 30 cm and mass 1.5 kg. Neglect the mass of the axle and sprocket. If a force of

22 N is applied tangentially to a sprocket of radius 6 cm for 6 s, what angular speed does the wheel achieve, assuming it rolls without slipping?

1 answer:

vredina [299]3 years ago

7 0

Answer:

The angular speed after 6s is $\omega = 1466.67s^{-1}$ .

Explanation:

The equation

$I\alpha = Fd$

relates the moment of inertia $I$ of a rigid body, and its angular acceleration $\alpha$ , with the force applied $F$ at a distance $d$ from the axis of rotation.

In our case, the force applied is $F = 22N$ , at a distance $d = 6cm =0.06m$ , to a ring with the moment of inertia of $I =mr^2$ ; therefore, the angular acceleration is

$\alpha =\frac{Fd}{I}$

$\alpha =\frac{22N*0.06m}{(1.5kg)*(0.06)^2}$

$\alpha = 244.44\: s^{-2}$

Therefore, the angular speed $\omega$ which is

$\omega = \alpha t$

after 6 seconds is

$\omega = 244.44$\: s^{-2}* 6s$

$\boxed{\omega = 1466.67s^{-1}}$

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A single-turn circular loop of radius 6 cm is to produce a field at its center that will just cancel the earth's field of magnit

djverab [1.8K]

Answer:

The current is $I = 6.68 \ A$

Explanation:

From the question we are told that

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The earth's magnetic field is $B_e = 0.7G= 0.7 G * \frac{1*10^{-4} T}{1 G} = 0.7 *10^{-4} T$

The number of turns is $N =1$

Generally the magnetic field generated by the current in the loop is mathematically represented as

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Now for the earth's magnetic field to be canceled out the magnetic field generated by the loop must be equal to the magnetic field out the earth

$B = B_e$

=> $B_e = \frac{\mu_o * N * I }{ 2 * r}$

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3 years ago

on aircraft carriers, catapults are used to accelerate jet air craft to flight speeds in a short distance. One such catapult tak

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Acceleration = (change in speed) / (time for the change)

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The jet's change in speed = (70 m/s) - (zero) = 70 m/s

So acceleration = (70 m/s) / (2.5 s)

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That's about 113,000 pounds ! Maybe the part of the airplane that the catapult pushes on can't handle any more force than that. Or maybe that's the most force the catapult can deliver.

Also, the REACTION force on the catapult is the same 113,000 pounds. Maybe the hooks or the chains or the struts on the catapult can't handle any more force than that.

That's almost 57 tons for gosh sakes ! Maybe the DECK of the carrier can't handle more force than that, and that's why they can't launch the airplane with acceleration of more than 2.9 G's .

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3 years ago

A 40 kg slab rests on a frictionless floor. A 10 kg block rests on top of the slab (Fig. 6-58). The coefficient of static fricti

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Answer:

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Answer:

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R = mg - ma

R measures its apparent weight . Spring scale will measure his apparent weight.

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4 years ago