Ask question

Ask question

All categories

4 years ago

11

When it is at rotating at full speed, a disk drive in a certain old computer game system revolves once every 0.050 seconds. Star

ting from rest, it takes two revolutions for the disk to reach full speed. Assuming that the angular acceleration of the disk is constant, what is its angular acceleration while it is speeding up

1 answer:

Gemiola [76]4 years ago

3 0

Answer:

α = 200*π rad/sec²

Explanation:

If the disk drive revolves once every 0.05 sec, this means that the angular velocity can be expressed as follows:

$\omega = \frac{2*\pi}{0.05s} = 40*\pi rad/sec$

(just applying the definition of angular velocity)
Now, it the disk started from rest, and took two revolutions to reach to full speed, assuming that the angular acceleration is constant, we can find the angular acceleration applying this kinematic equation:

$\omega_{f} ^{2} -\omega_{0} ^{2} = 2*\alpha * \Delta\theta$

where Δθ, is the angle rotated while it went from rest to full speed, i.e., 2 revolutions, or 2*π rads.
Replacing by the value that we found for ωf (as ω₀ = 0), we can solve for α :

$\alpha =\frac{\omega_{f}^{2} }{2*\Delta\theta} = \frac{(40*\pi)^{2} }{2*2*2*\pi } =\\ \\ \alpha = 200*\pi rad/sec2$

You might be interested in

Suppose you're on a hot air balloon ride, carrying a buzzer that emits a sound of frequency f. If you accidentally drop the buzz

amid [387]

Answer:

(C) The frequency decrease and intensity decrease

Explanation:

The Doppler effect describes the change in frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source, or the wave source is moving relative to the observer, or both.

if the observer and the source move away from each other as is the case for this problem, the wavelength heard by the observer is bigger.

The frequency is the inverse from the wavelength, so the frequency heard will increase.

The sound intensity depends inversely on the area in which the sound propagates. When the buzzer is close, the area is from a small sphere, but as the buzzer moves further away, the wave area will be from a larger sphere and therefore the intensity will decrease.

7 0

4 years ago

Which statement is true for a car that first goes around a curve of radius r at a constant speed v, and then goes around the sam

sasho [114]

Centripetal force is equal to (mv^2)/r
The way I use to answer these question is to set every variable to 1
m=1
v=1
r=1
so centripetal force =1
then change the variable we're looking at
and since we're find when it's half we could either change it to 1/2 or 2, but 2 is easier to use
m=1
v=2
r=1
((1)×(2)^2)/1=4
So the velocity in the 1st part is half the velocity in the 2nd part and the centripetal force is 4× less
The answer is the centripetal force is 1/4 as big the second time around

3 0

4 years ago

A rocket is launched at an angle of 53.0° above the horizontal with an initial speed of 103 m/s. The rocket moves for 3.00 s alo

Serggg [28]

Before the engines fail $(0\le t\le3.00\,\rm s)$ , the rocket's horizontal and vertical position in the air are

$x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t^2$

$y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t^2$

and its velocity vector has components

$v_x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t$

$v_y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t$

After $t=3.00\,\rm s$ , its position is

$x=273\,\rm m$

$y=362\,\rm m$

and the rocket's velocity vector has horizontal and vertical components

$v_x=120\,\frac{\rm m}{\rm s}$

$v_y=159\,\frac{\rm m}{\rm s}$

After the engine failure $(t>3.00\,\rm s)$ , the rocket is in freefall and its position is given by

$x=273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)t$

$y=362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

and its velocity vector's components are

$v_x=120\,\frac{\rm m}{\rm s}$

$v_y=159\,\frac{\rm m}{\rm s}-gt$

where we take $g=9.80\,\frac{\rm m}{\mathrm s^2}$ .

a. The maximum altitude occurs at the point during which $v_y=0$ :

$159\,\frac{\rm m}{\rm s}-gt=0\implies t=16.2\,\rm s$

At this point, the rocket has an altitude of

$362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)(16.2\,\rm s)-\dfrac g2(16.2\,\rm s)^2=1650\,\rm m$

b. The rocket will eventually fall to the ground at some point after its engines fail. We solve $y=0$ for $t$ , then add 3 seconds to this time:

$362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=34.6\,\rm s$

So the rocket stays in the air for a total of $37.6\,\rm s$ .

c. After the engine failure, the rocket traveled for about 34.6 seconds, so we evalute $x$ for this time $t$ :

$273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)(34.6\,\rm s)=4410\,\rm m$

5 0

3 years ago

"A ball with a mass of 0.1 kg is rolling at a velocity of 5 m/s. What is its

Mumz [18]

Answer:

velocity

Explanation:

because the si unit of mass is kg, velocity is m/s, acceleration is m/S2 , moment is kgm2/s . so 5 is given as velocity.

3 0

2 years ago

What is the equivalent resistance of a circuit that contains three 10.0 12

lana66690 [7]

Answer:

Explanation:

The equivalent resistance for three resistors connected in parallel is given as

(1/R)=(1/R₁)+(1/R₂)+(1/R₃)

now we.need to.insert the value of 3 resistances but only 2 are given in the question.

3 0

3 years ago

Read 2 more answers