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Digiron [165]
2 years ago
12

Three solid, uniform, cylindrical flywheels, each of mass 65.0 kg and radius 1.47 m, rotate independently around a common axis t

hrough their centers. Two of the flywheels rotate in one direction at 8.94 rad/s, but the other one rotates in the opposite direction at 3.42 rad/s.
Required:
Calculate the magnitude of the net angular momentum of the system.
Physics
1 answer:
Mkey [24]2 years ago
3 0

Answer:

the angular momentum is 1015.52 kg m²/s

Explanation:

given data

mass of each flywheel, m = 65 kg

radius of flywheel, r = 1.47 m

ω1 = 8.94 rad/s

ω2 = - 3.42 rad/s

to find out

magnitude of the net angular momentum

solution

we get here Moment of inertia that is express as

I = 0.5 m r²    .................1

put here value and we get

I = 0.5 × 65 × 1.47 × 1.47

I = 70.23 kg m²

and

now we get here Angular momentum that is express as

L = I × ω    ...........................2

and Net angular momentum will be

L = 2 × I x ω1 - I × ω2

put here value and we get

L = 2 × 70.23 × 8.94 - 70.23 × 3.42

L = 1015.52 kg m²/s

so

the angular momentum is 1015.52 kg m²/s

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lidiya [134]

Answer:

EMF = 316 Volts

Explanation:

As we know that magnetic flux through the coil is given by

\phi = NBA

now by Faraday law we know that rate of change in magnetic flux is equal to the EMF induced in the coil

so we have

-\frac{d\phi}{dt} = EMF

EMF = -NA\frac{dB}{dt}

now we have

B = 50 sin(10\pi t)

A = \pi r^2 = \pi(0.04)^2 = 5.03 \times 10^{-3} m^2

now we have

EMF = -(40)(5.03 \times 10^{-3})\frac{d(50 sin(10\pi t))}{dt}

EMF = -0.2012(50 \times 10\pi)cos(10\pi t)

EMF = -316 cos(10\pi t)

now at t = 0.10 s

EMF = 316 Volts

3 0
2 years ago
Suppose you are talking by interplanetary telephone to your friend, who lives on the Moon. He tells you that he has just won a n
Savatey [412]

Answer:

The friend on moon will be richer.

Explanation:

We must calculate the mass of gold won by each person, to tell who is richer. For that purpose we will use the following formula:

W = mg

m = W/g

where,

m = mass of gold

W = weight of gold

g = acceleration due to gravity on that planet

<u>FOR FRIEND ON MOON</u>:

W = 1 N

g = 1.625 m/s²

Therefore,

m = (1 N)/(1.625 m/s²)

m(moon) = 0.6 kg

<u>FOR ME ON EARTH</u>:

W = 1 N

g = 9.8 m/s²

Therefore,

m = (1 N)/(9.8 m/s²)

m(earth) = 0.1 kg

Since, the mass of gold on moon is greater than the mass of moon on earth.

<u>Therefore, the friend on moon will be richer.</u>

7 0
2 years ago
1. Two identical bowling balls of mass M and radius R roll side by side at speed v0 along a flat surface. Ball 1 encounters a ra
UNO [17]

Answer:

1/2

Explanation:

We need to make a couple of considerations but basically the problem is solved through the conservation of energy.  

I attached a diagram for the two surfaces and begin to make the necessary considerations.

Rough Surface,

We know that force is equal to,

F_r = mgsin\theta

F_r = \mu N

F_r = \mu mg cos\theta

Matching the two equation we have,

\mu N = \mu mg cos\theta

\mu = tan\theta

Applying energy conservation,

\frac{1}{2}mv^2_0+\frac{1}{2}I_w^2 = F_r*d+mgh_1

\frac{1}{2}mv^2_0+\frac{2}{5}mR^2\frac{V_0^2}{R^2} = F_r*d+mgh_1

\frac{1}{2}mv^2_0+\frac{mv_0^2}{5} = mgsin\theta \frac{h_1sin\theta}+mgh_1

\frac{v_0^2}{2}+\frac{v_0^2}{5} = gh_1+gh_1

h_1 = \frac{1}{2g}(\frac{v_0^2}{2}+\frac{v_0^2}{5})

Frictionless surface

\frac{1}{2}mv_0^2+\frac{1}{2}I\omega^2 = mgh_2

\frac{1}{2}m_v^2+\frac{1}{2}\frac{2}{5}mR^2\frac{v_0^2}{R^2} =mgh_2

\frac{v_0^2}{2}+\frac{v_0^2}{5} = gh_2

h_2 = \frac{1}{g}(\frac{V_0^2}{2}+\frac{v_0^2}{5})

Given the description we apply energy conservation taking into account the inertia of a sphere. Then the relation between h_1 and h_2 is given by

\frac{h_1}{h_2} = \frac{\frac{1}{2g}(\frac{v_0^2}{2}+\frac{v_0^2}{5})}{\frac{1}{g}(\frac{V_0^2}{2}+\frac{v_0^2}{5})}

\frac{h_1}{h_2} = \frac{1}{2}

8 0
2 years ago
An 888.0 kg elevator is moving downward with a velocity of 0.800 m/s. It decelerates uniformly and comes to a stop in a distance
bagirrra123 [75]

Answer:

The value of tension on the cable T = 1065.6 N

Explanation:

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Initial velocity ( u )= 0.8 \frac{m}{sec}

Final velocity ( V ) = 0

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Now use third law of motion V^{2} = u^{2} - 2 a s

Put all the values in above formula we get,

⇒ 0 = 0.8^{2} - 2 × a ×0.2667

⇒ a = 1.2 \frac{m}{sec^{2} }

This is the deceleration of the box.

Tension in the cable is given by T = F = m × a

Put all the values in above formula we get,

T = 888 × 1.2

T = 1065.6 N

This is the value of tension on the cable.

5 0
2 years ago
A hummingbird can a flutter its wings 4800 times per minute what is the frequency of wing flutters per second
Scilla [17]

the answer i got is 80 i hope this helpss!!!!

7 0
2 years ago
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