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lesya [120]
3 years ago
9

A 4.2-m-diameter merry-go-round is rotating freely with an angular velocity of 0.79 rad/s . Its total moment of inertia is 1790

kg⋅m2 . Four people standing on the ground, each of mass 70 kg , suddenly step onto the edge of the merry-go-round.
Physics
1 answer:
Nezavi [6.7K]3 years ago
8 0

Answer:

w_2=0.467rad/s

Explanation:

Four people standing on the ground each of mass and usually this questions have to find the final angular velocity

m_t=4*70kg=280kg

The radius r=4.2/2=2.1m

Angular velocity  w_1=0.79rad/s

The moment of inertia total is I_t=1790 kg/m^2

Momento if inertia

I_1=m_t*r^2

I_1=280kg*(2.1m)^2=1234.8kg*m^2

Angular momentum

I_1*w_1=I_t*w_2

Solve to w2

w_2=\frac{I_1*w_1}{I_t}

w_2=\frac{1790kg*m^2*0.79rad/s}{3024.8kg*m^2}

w_2=0.467rad/s

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Correct question:

A solenoid of length 0.35 m and diameter 0.040 m carries a current of 5.0 A through its windings. If the magnetic field in the center of the solenoid is 2.8 x 10⁻² T, what is the number of turns per meter for this solenoid?

Answer:

the number of turns per meter for the solenoid is 4.5 x 10³ turns/m.

Explanation:

Given;

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The number of turns per meter for the solenoid is calculated as follows;

B =\mu_o  I(\frac{N}{L} )\\\\B =  \mu_o  I(n)\\\\n = \frac{B}{\mu_o I} \\\\n = \frac{2.8 \times 10^{-2}}{4 \pi \times 10^{-7} \times 5.0} \\\\n = 4.5 \times 10^3 \ turns/m

Therefore, the number of turns per meter for the solenoid is 4.5 x 10³ turns/m.

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3 years ago
A truck travels up a hill with a 5.7° incline. The truck has a constant speed of 22 m/s. What is the horizontal component of the
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Answer:

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Please see the attached figure for a better understanding of the problem.

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To find vx, let´s use the following trigonometric rule of right triangles:

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cos 5.7° = vx / 22 m/s

22 m/s · cos 5.7° = vx

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Explanation: Hope it helps you:))))

have a good day

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The question is incomplete. I can help you by adding the information missing. They want you to calculate a) the radius of the cyclotron orbit for an electron with speed 1.0 * 10^6 m/s^2 and b) the radius of a cyclotron orbit for a proton with speed 5.0 * 10^4 m/s.

The two tasks involve combining the equations of the magnectic force and the centripetal force in a circular motion.

When you do that, you will obtain an expression to find the radius of the circular motion, which is the radius of the cyclotron that impulses the particles.

a)

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q is the charge of the electron = 1.6 * 10^ -19 C
v is the speed = 1.0 * 10 ^ 6 m/s
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Centripetal force, F = m*Ac = m * v^2 / R

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Ac = centripetal acceleration
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Now equal the two forces: q*v*B = m * v^2 / R => R =  m*v / (q*B)

=> R = (9.11 * 10^31 kg) (1.0*10^6m/s) / [ (1.6 * 10^-19C)* (5.0 * 10^-5T) ]

=> R = 0.114 m

b) The equations are the same, just now use the speed, charge and mass of the proton instead of those of the electron.

R = m*v / (qB) = (1.66*10^-27 kg)(5.0*10^4 m/s) / [(1.6*10^-19C)(5*10^-5T)]

=> R = 10.4 m

 

4 0
3 years ago
At the accounting break-even point, Swiss Mountain Gear sells 22,940 ski masks at a price of $19 each. At this level of producti
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Answer:

fixed cost $231220

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