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Nutka1998 [239]

2 years ago

15

You toss a conductive open ring of diameter d = 1.75 cm up in the air. The ring is flipping around a horizontal axis at a rate o

f 6.25 flips per second. One flip is a full rotation. At your location, the earth's magnetic field is B e = 30.5 μT. What is the maximum emf induced in the ring?

1 answer:

Mama L [17]2 years ago

8 0

Answer:

The maximum emf induced in the ring

= (2.882 × 10⁻⁷) V

Explanation:

According to the law of electromagnetic induction, the emf induced in the ring is given by

E = N BA w sin wt

The maximum emf induced is

E = N BA w

B = 30.5 μT = (30.5 × 10⁻⁶) T

A = (πD²/4)

D = 1.75 cm = 0.0175 m

A = (π×0.0175²/4) = 0.000240625 m²

Nw = 2π × 6.25 = 39.29 rad/s

E = 30.5 × 10⁻⁶ × 0.000240625 × 39.29

E = (2.882 × 10⁻⁷) V

Hope this Helps!!!

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The angle θ is slowly increased. Write an expression for the angle at which the block begins to move in terms of μs.

Reika [66]

Answer:

$tan \theta = \mu_s$

Explanation:

An object is at rest along a slope if the net force acting on it is zero. The equation of the forces along the direction parallel to the slope is:

$mg sin \theta - \mu_s R =0$ (1)

where

$mg sin \theta$ is the component of the weight parallel to the slope, with m being the mass of the object, g the acceleration of gravity, $\theta$ the angle of the slope

$\mu_s R$ is the frictional force, with $\mu_s$ being the coefficient of friction and R the normal reaction of the incline

The equation of the forces along the direction perpendicular to the slope is

$R-mg cos \theta = 0$

where

R is the normal reaction

$mg cos \theta$ is the component of the weight perpendicular to the slope

Solving for R,

$R=mg cos \theta$

And substituting into (1)

$mg sin \theta - \mu_s mg cos \theta = 0$

Re-arranging the equation,

$sin \theta = \mu_s cos \theta\\\rightarrow tan \theta = \mu_s$

This the condition at which the equilibrium holds: when the tangent of the angle becomes larger than the value of $\mu_s$ , the force of friction is no longer able to balance the component of the weight parallel to the slope, and so the object starts sliding down.

4 0

2 years ago

A 36,287 kg truck has a momentum of 907,175 kg • m/s. What is the trucks velocity.

Airida [17]

Momentum = Mass x Velocity

Put the values where they belong and solve for Velocity.

In this case, since Mass is being multiplied by Velocity, to solve for be Velocity you would divide both sides by Velocity. The velocity will equal the momentum divided by the mass.

4 0

2 years ago

The magnetic field of an electromagnetic wave in a vacuum is Bz =(2.4μT)sin((1.05×107)x−ωt), where x is in m and t is in s. You

tatiyna

Answer:

Explanation:

Given

$B_z=(2.4\mu T)\sin (1.05\times 10^7x-\omega t)$

Em wave is in the form of

$B=B_0\sin (kx-\omega t)$

where $\omega =frequency\ of\ oscillation$

$k=wave\ constant$

$B_0=Maximum\ value\ of\ Magnetic\ Field$

Wave constant for EM wave k is

$k=1.05\times 10^7 m^{-1}$

Wavelength of wave $\lambda =\frac{2\pi }{k}$

$\lambda =\frac{2\pi }{1.05\times 10^7}$

$\lambda =5.98\times 10^{-7} m$

7 0

2 years ago

A closely wound rectangular coil of 80 turns has dimensions 25.0 \rm cm by 40.0 \rm cm. The plane of the coil is rotated from a

Pepsi [2]

Answer:

88.3

Explanation:

Emf in a rotating coil is given by rate of change of flux:

E= dФ/dt=(NABcos∅)/ dt

N: number of turns in the coil= 80

A: area of the coil= 0.25×0.40= 0.1

B: magnetic field strength= 1.1

Ф: angle of rotation= 90- 37= 53

dt= 0.06s

E= (80 × 0.4× 0.25×1.10 × cos53)/0.06= 88.3V

4 0

2 years ago

Read 2 more answers

A girl and a boy are riding on a merry-go-round that is turning at a constant rate. The girl is near the outer edge, and the boy

skad [1K]

(a) Both the girl and the boy have the same nonzero angular displacement.

Explanation:

The angular displacement of an object moving in uniform circular motion, as the boy and the girl on the merry-go-round, is given by

$\theta= \omega t$

where

$\omega$ is the angular speed

t is the time interval

For a uniform object in uniform circular motion, all the points of the object have same angular speed. This means that the value of $\omega$ is the same for the boy and the girl.

Therefore, if we consider the same time interval t, the boy and the girl will also have same nonzero angular displacement.

(b) The girl has greater linear speed.

Explanation:

The linear (tangential) speed of a point along the merry-go-round is given by

$v=\omega r$

where

$\omega$ is the angular speed

r is the distance of the point from the centre of the merry-go-round

In this problem, the girl is near the outer edge, while the boy is closer to the centre: since the value of $\omega$ is the same for both, this means that the value of r is larger for the girl, so the girl will also have a greater linear speed.

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2 years ago