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Makovka662 [10]
3 years ago
14

A 200 turn coil is in a uniform magnetic field that is decreasing at the rate 0.20 T/s. The coil is perpendicular to the field a

nd its dimensions are 0.20 m by 0.40 m. What is the magnitude of the induced emf in the coil
Physics
1 answer:
sattari [20]3 years ago
8 0

Answer:

emf = 3.2V

Explanation:

In order to calculate the magnitude of the induced emf in the coil you use the following formula:

emf=-N\frac{d\Phi_B}{dt}       (1)

N: turns of the coil = 200

ФB: magnetic flux = A*B

A: area of the coil = (0.20m)(0.40m) = 0.08m²

B: magnitude of the magnetic field

You take into account that the area of the coil is constant, while magnetic field changes on time. Then, the equation (1) becomes:

emf=-NA\frac{dB}{dt}         (2)

dB/dt =  rate of change of the magnetic field = -0.20T/s (it is decreasing)

You replace the values of all parameters in the equation (2):

emf=-200(0.08m^2)(-0.20T/s)=3.2V

The induced emf in the coil is 3.2V

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Monochromatic light of variable wavelength is incident normally on a thin sheet of plastic film in air. The reflected light is a
BartSMP [9]

Answer:

thickness t = 528.433 nm

Explanation:

given data

wavelength λ1 = 477.1 nm

wavelength λ2 = 668.0 nm

n = 1.58

solution

we know for constructive interference condition will be

2 × t × μ = (m1+0.5) × λ1     ....................1

2 × t × μ = (m2+0.5) × λ2     ....................2

so we can say from equation 1 and 2

(m1+0.5) × λ1 = (m2+0.5) × λ2

so

\frac{\lambda 2}{\lambda 1} = \frac{m1+0.5}{m2+0.5}     ..............3

put here value and we get  

\frac{668.0}{477.1} = \frac{m1+0.5}{m2+0.5}  

\frac{m1+0.5}{m2+0.5}   = 1.4

\frac{m1+0.5}{m2+0.5}  = \frac{7}{5}   ...................4

so we here from equation 4

m1+0.5  = 7

m1 = 3    .................5

m2+0.5 = 4

m2 = 2    .................6

so now put value in equation  1

2 × t × μ = (m1+0.5) × λ1

2 × t × 1.58 = (3+0.5) ×  477.1

solve it we get

thickness t = 528.433 nm

4 0
3 years ago
The diagram represents water waves travelling from deep water into an area of much shallower water.
Agata [3.3K]
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;kk

5 0
3 years ago
A baseball is hit almost straight up into the air with a speed of 26 m/s . Estimate how high it goes.
Natalka [10]

Answer:

The maximum height of the ball is 34.5 m.

The ball is 5.31 s in the air.

Explanation:

Hi there!

The equations for the height and velocity of the baseball that is hit straight up are as follows:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height of the baseball at time t.

y0 = initial height.

v0 = initial velocity.

t = time.

g =  acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity at time t.

If we place the origin of the frame of reference at the place where the baseball is hit, then, y0 = 0.

To calculate how high it goes, we have to obtain the time at which the ball is at maximum height. At that point, the velocity is 0. Then using the equation of velocity:

v = v0 + g · t

0 = 26 m/s - 9.8 m/s² · t

-26 m/s / -9.8 m/s² = t

t = 2.65 s

The height at that time will be the maximum height:

y = y0 + v0 · t + 1/2 · g · t²        (y0 = 0)

y = 26 m/s · 2.65 s - 1/2 · 9.8 m/s² · (2.65 s)²

y = 34.5 m

The maximum height of the ball is 34.5 m

If it takes the ball 2.65 s to reach the maximum height it will take another 2.65 s to return to the initial position. Then, the time it will be in the air is (2.65 s + 2.65 s) 5.30 s. However, let´s calculate the time it takes the ball to reach the initial position using the equation for height.

At the initial position y = 0. Then:

y = y0 + v0 · t + 1/2 · g · t²        (y0 = 0)

0 = 26 m/s · t - 1/2 · 9.8 m/s² · t²

0 = t (26 m/s - 1/2 · 9.8 m/s² · t)      (t = 0 when the ball is hit)

0 = 26 m/s - 1/2 · 9.8 m/s² · t

-26 / -4.9 m/s² = t

t = 5.31 s     ( the difference with the 5.30 s obtained above is due to rounding the time to 2.65 s).

The ball is 5.31 s in the air.

Have a nice day!

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