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2 years ago

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Imagine that you are a circus performer riding a uni-cycle (forwards) across the stage. What is the direction of the angular vel

ocity of the single wheel?

1 answer:

Sergio039 [100]2 years ago

3 0

The direction of the angular velocity of the single wheel is the clockwise direction

<h3>Angular velocity</h3>

The angular velocity is the number of revolutions per second of the single wheel. It is given by ω = v/r where

v = linear speed of the wheel and
r = radius of the wheel.

<h3>Direction of the angular velocity of the single wheel</h3>

Since the wheels moves forward and the velocity of the wheel is always tangential to the wheel causing it to move forward, its angular velocity is in the clockwise direction

So, the direction of the angular velocity of the single wheel is the clockwise direction

Learn more about angular velocity here:

brainly.com/question/20912191

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Suppose you are on an airplane moving at high speed. If you flip a coin straight up it will land in your lap rather than a great

maxonik [38]

It will land in your lap because there's different frames of motion relative to yourself. For example, if you're running at a speed of 6 mph, it doesn't mean you'll run as fast as the Earth spins. Also, since you're on the interior of the plane, any kind of wind or weather on the outside will not affect the coin. A law to back up this claim is Einsteins Special Law of Relativity.

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4 years ago

Read 2 more answers

What is free fall? Explain it in brief.

xxTIMURxx [149]

· free fall is any motion of a body where gravity is the only forceacting upon it. In the context of general relativity, where gravitation is reduced to a space-time curvature, a body in free fall has no force acting on it.

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3 years ago

A box is pushed 40 m by a mover. The amount of work done was 2,240 j. How much force was exerted on the box

Georgia [21]

The force exerted on the box is 56 N

Explanation:

The work done by a force on an object is given by

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the object

$\theta$ is the angle between the direction of the force and of the displacement

For the box in this problem, we have:

W = 2240 J is the work done

d = 40 m is the displacement of the box

Assuming that the force is parallel to the displacement, $\theta=0$

Solving the equation for F, we find the force exerted on the box:

$F=\frac{W}{d cos \theta}=\frac{2240}{(40)(cos 0)}=56 N$

Learn more about work:

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brainly.com/question/6443626

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3 years ago

When we look up in the sky the Sun appears about as big as the moon; however, we know that the Sun is much further away. Given t

Nady [450]

Answer: well since you can fit one million Earths in the sun just multiply.

hope this helps happy thanksgiving ;)

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3 years ago

The tape in a videotape cassette has a total

xxMikexx [17]

Answer:

$w=19.76 \ rad/s$

Explanation:

<u>Circular Motion </u>

Suppose there is an object describing a circle of radius r around a fixed point. If the relation between the angle of rotation by the time taken is constant, then the angular speed is also constant. If that relation increases or decreases at a constant rate, the angular speed is given by:

$w=w_o+\alpha \ t$

Where $\alpha$ is the angular acceleration and t is the time. If the object was instantly released from the circular path, it would have a tangent speed of:

$\displaystyle v_t=w.r$

We have two reels: one loaded with the tape to play and the other one empty and starting to fill with tape. They both rotate at different angular speeds, one is increasing and the other is decreasing as the tape goes from one to the other. We'll assume the tangent speed is constant for both (so the tape can play correctly). Let's call $w_1$ the angular speed of the loaded reel and $w_2$ that from the empty reel. We have

$w_1=w_{o1}+\alpha_1 \ t$

$w_2=w_{o2}+\alpha_2 \ t$

If $r_f=35\ mm=0.035\ m$ is the radius of the reel when it's full of tape, the angular speed for the loaded reel is computed by

$\displaystyle w_{01}=\frac{v_t}{r_f}$

The tangent speed is computed by knowing the length of the tape and the time needed to fully play it.

$t=1.8\ h=1.8*3600=6480\ sec$

$\displaystyle v_t=\frac{x}{t}=\frac{249\ m}{6480\ sec}=0.0384 \ m/s$

$\displaystyle w_{01}=\frac{0.0384}{0.035}=1,098 \ rad/s$

If $r_e=10\ mm=0.001\ m$ is the radius of the reel when it's empty, the angular speed for the empty reel is computed by

$\displaystyle w_{02}=\frac{0.0384}{0.001}=38.426 \ rad/s$

The full reel goes from $w_{01}$ to $w_{02}$ in 6480 seconds, so we can compute the angular acceleration:

$\displaystyle \alpha_1=\frac{w_{02}-w_{01}}{6480}$

$\displaystyle \alpha_1=\frac{38.426-1.098}{6480}=0.00576 \ rad/sec^2$

The empty reel goes from $w_{02}$ to $w_{01}$ in 6480 seconds, so we can compute the angular acceleration:

$\displaystyle \alpha_2=\frac{1.098-38.426}{6480}=-0.00576 \ rad/sec^2$

So the equations for both reels are

$w_1=1.098+0.00576 \ t$

$w_2=38.426-0.00576 \ t$

They will be the same when

$1.098+0.00576 \ t=38.426-0.00576 \ t$

Solving for t

$\displaystyle t=\frac{38.426-1.098}{0.0115}$

$t=3240 \ sec$

The common angular speed is

$w_1=1.098+0.00576 \ 3240=19.76 \ rad/s$

$w_2=38.426-0.00576 \ 3240=19.76 \ rad/s$

They both result in the same, as expected

$\boxed{w=19.76 \ rad/s}$

6 0

4 years ago