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jarptica [38.1K]

3 years ago

11

Two gerbils run in place with a linear speed of 0.50 m/s on an exercise wheel that is shaped like a hoop. Find the angular momen

tum of the system if each gerbil has a mass of 0.22 kg and the exercise wheel has a radius of 8 cm and a mass of 4.5 g.

1 answer:

Brilliant_brown [7]3 years ago

7 0

Answer:

$L = 0.018 kg m^2/s$

Explanation:

Angular speed and linear speed is related to each other as

$v = R\omega$

here we know that

$v = 0.50 m/s$

$R = 0.08 m$

now we have

$\omega = \frac{v}{R}$

$\omega = \frac{0.50}{0.08}$

$\omega = 6.25 rad/s$

Now we know that moment of inertia of the system is given as

$I = (2M + m)R^2 [tex][tex]I = (2\times 0.22 + 0.0045)(0.08)^2$

$I = 2.84 \times 10^{-3} kg m^2$

Now angular momentum of the system is given as

$L = I\omega$

$L = (2.84 \times 10^{-3})(6.25)$

$L = 0.018 kg m^2/s$

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The earth's tectonic plates are directly propelled by the convection in the __________.

frozen [14]

The missing word here is <u>Asthenosphere.</u><u> </u>

The convection in the asthenosphere directly propels the tectonic plates of the earth.

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The older, denser portions of the lithosphere that are dragged downward in subduction zones are stored in the asthenosphere, according to the theory of plate tectonics.

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3 0

1 year ago

A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate

Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by $b$ ) is at 36 ft. from the wall is the following:

$\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s$

Explanation:

The Area of the triangle is given by $A=h\times b$ where $h=\sqrt{l^2-b^2}$ (by using the Pythagoras' Theorem) and $b$ is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

$A=\sqrt{l^2-b^2}b$

The rate of change of the area is given by its time derivative

$\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)$

$\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}$

$\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Product rule

$\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Chain rule

$\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$

$\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)$

In here we can identify $b=36\, ft$ , $l=39$ and $\frac{db}{dt}=8\,ft/s$ .

The result is then

$\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s$

3 0

2 years ago

A cyclotron is to accelerate protons to an energy of 5.4 MeV. The superconduction electromagnet of the cyclotron produces a 2.9T

mart [117]

<h3><u>Answer;</u></h3>

Radius = 0.0818 m

Angular velocity = 2.775 × 10^7 rad/sec

<h3><u>Explanation;</u></h3>

The mass of proton m=1.6748 × 10^-27 kg;

Charge of electron e= 1.602 × 10^-19 C;

kinetic energy E= 2.7 MeV

= 2.7 × 10^6 × 1.602 × 10^-19 J;

= 4.32 × 10^-13 Joules

But; K.E =0.5m*v^2,

Hence v=√(2K.E/m)

Velocity = 2.27 × 10^7 m/s

Angular velocity, ω = v/r

Therefore; V = ωr

Hence; V = √(2K.E/m) = ωr

r= √(2E/m)/w = √E*√(2*m)/(eB)

= √E * √(2×1.6748×10^-27)/(1.602×10^-19 ×2.9)

but E = 4.32 × 10^-13 Joules

r = 0.0818 m

Angular speed

Angular velocity, ω = v/r , where r is the radius and v is the velocity

Therefore;

Angular velocity = 2.27 × 10^7 / 0.0818 m

= 2.775 × 10^7 rad /sec

3 0

2 years ago

Help me answer this ASAP

olga_2 [115]

Answer:

The correct answer is Option A.

7 0

3 years ago

A roller coaster car is loaded with passengers and has a mass of 500 kg along with a speed of 18 meters/second at the dip. The r

GenaCL600 [577]

The roller coaster is moving in a circular path, so the force that must be computed is the centripetal force. The centripetal force is the force acting on an object undergoing circular motion and is directed towards the center of the circular path. This is computed using:
F = mv²/r
F = (500 * 18²) / 12
F = 13,500 N

Now, at the bottom of the track, the track is also supporting the weight of the car and its passengers, which is:
W = mg
W = 500 * 9.81
W = 4,905 N

The total reactive force exerted by the track to counter the centripetal force and the weight of the car is:
F = 13,500 + 4,905
F = 18,405 Newtons

7 0

2 years ago

Read 2 more answers