Physics, 15.04.2020 15:36 momoney5746

A 250 g toy car is placed on a narrow 60-cm-diameter track with wheel grooves that keep the car going in a circle. The 1.5 kg track is free to turn on a frictionless, vertical axis. The spokes have negligible mass. After the car's switch is turned on, it soon reaches a steady speed of 0.74 m/s relative to the track. What then is the track's angular velocity, in rpm?

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Answer from: shrafe

$\omega_{t}=-3.9 rpm$

Explanation:

Let's use the conservation of angular momentum here. If there is a conservation the addition of both must be zero.

$L_{c}+L_{t}=0$

The momentum of inertia of both are:

$I_{c}=m_{c}R^{2}$

$I_{t}=m_{t}R^{2}$

The angular momentum is the product between angular velocity and momentum of inertia.

$I_{c}\omega_{c}+I_{t}\omega_{t}=0$

Let's solve it for ω(t).

$\omega_{t}=-\frac{I_{c}\omega_{c}}{I_{t}}$

$\omega_{t}=-\frac{m_{c}R^{2}\omega_{c}}{m_{t}R^{2}}$

But $\omega=V/R$

R is the radius R=D/2 = 30 cm = 0.3 mV is the steady speed V = 0.74 m/s

$\omega_{t}=-\frac{m_{c}R^{2}*V/R}{m_{t}R^{2}}$

$\omega_{t}=-\frac{m_{c}V}{m_{t}R}$

$\omega_{t}=-\frac{0.25*0.74}{1.5*0.3}$

$\omega_{t}=-0.41 rad/s$

Knowing that 2π rad is a rev. We have.

$\omega_{t}=-3.9 rpm$

The minus sign means the track is moving opposite of the car.

I hope it helps you!

Answer from: litzy9506

the track's angular velocity = 3.363 rpm

Explanation:

Given that ;

A 250 g toy car is placed on a narrow 60-cm-diameter track

mass m of the car = 250 g = 0.25 kg

diameter of the car = 60 cm

radius of the car = 60/2 = 30 cm = 0.3 m

The 1.5 kg track is free to turn on a frictionless, vertical axis.

mass M of the track = 1.5 kg

steady speed v = 0.74 m/s

the track's angular velocity, in rpm =???

Now; the derived equation for the conservation of angular momentum can be expressed as:

$\omega = \frac{mv}{r(m+M)}$

$\omega = \frac{0.25*0.74}{0.3(0.25+1.5)}$

$\omega = 0.352 \ rad/ sec$

to rpm ; we have:

$\omega = 0.352 \ rad/ sec (\frac{60 s}{1 m})(\frac{1 rev}{2(3.14)rad})$

$\omega = 3.363 \ rpm$

Thus, the track's angular velocity = 3.363 rpm

Answer from: Quest

as we probably am aware that our atmosphere or air have various levels, each level has different gasses, and each gas has another porousness. at the point when the light originates from lower porousness to higher penetrability, it alters its course. with the goal that implies that when the light originates from the moon to the earth, it alters it's course a lot more occasions. it essentially implies that the state of the light piece of the moon is changed on the earth.

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Answer from: Quest

$p_{i}$ = initial pressure

$p_{f}$ = final pressure = 2 $p_{i}$

$v_{i}$ = initial volume

$v_{f}$ = final volume = $p_{i}$ /4

$t_{i}$ = initial temperature = 20 c = 20 + 273 = 293 k

$t_{f}$ = final temperature = ?

using the equation

$\frac{p_{i}v_{i}}{t_{i}}= \frac{p_{f}v_{f}}{t_{f}}$

$\frac{p_{i}v_{i}}{293} = \frac{(2p_{i}) (v_{i}/4)}{t_{f}}$

$\frac{1}{293} = \frac{0.5}{t_{f}}$

$t_{f}=10$

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