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velikii [3]
3 years ago
9

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

position in which it makes an angle of 37.0 degrees with a magnetic field of 1.10 \rm T to a position perpendicular to the field. The rotation takes 0.0600 \rm s. What is the average emf EMF induced in the coil?What is the magnitude of the average emf E induced as the coil is rotated?E= ________ V

Physics
2 answers:
Law Incorporation [45]3 years ago
5 0

Answer:

E =90.25V

Explanation:

The illustration and formula to use in in the attachment.

  • The Turn is N=80
  • The dimension is 25cm by 40cm,  A=0.10m^{2}
  • Magnetic field of B is 1.7 T
  • Angle is ∅=80°
  • and time is Δt=0.0600 s

E_{av} = \frac{N*B*A IcosTita_{f}-cosTita_{i}  }{Dt}

      =\frac{80*1.70*0.10(cos(0)-cos(53)}{0.0600}

     =\frac{5.415}{0.0600}

E_{av} = 90.25V

Pepsi [2]3 years ago
4 0

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

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3 years ago
A 2.0 m conductor is formed into a square and placed in the horizontal A magnetic field is oriented above the horizontal with a
pashok25 [27]

Answer:

<h3>4.0Wb/m²</h3>

Explanation:

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8 0
3 years ago
You illuminate a slit with a width of 70.3 μm with a light of wavelength 719 nm and observe the resulting diffraction pattern on
Zielflug [23.3K]

Answer:

4.3 cm

Explanation:

We are given that

Width,d=70.3\mu m=70.3\times10^{-6} m

1\mu m=10^{-6} m

Wavelength,\lambda=719 nm=719\times 10^{-9} m

1nm=10^{-9} m

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sin\theta_{min}\approx \theta=\frac{\lambda}{d}=\frac{719\times 10^{-9}}{70.3\times 10^{-6}}=0.0102 rad

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8 0
3 years ago
A rock is dropped (from rest) off a bridge over the Merrimack River. The falling rock
rewona [7]

Answer:

31.25 meters or ~31 meters approximately

Explanation:

Let's see which of the 5 variables we are given since this is a constant acceleration problem.

  • v_i  \ \ \ \ \ \  t \\ v_f \ \ \ \ \ \triangle x \\ a

We want to find the height of the bridge, aka the vertical displacement of the rock. Let's set the upwards direction to be positive and the downwards direction to be negative.

We are told that the acceleration is 10 m/s² downward, so we have a = -10 m/s².

We are also told that the time it takes the rock to hit the water is 2.5 seconds. Time is the same regardless of the x- or y- direction, so we can say that t = 2.5 seconds.

Now, we aren't told this directly, but we can figure out that the velocity in the y-direction is 0 m/s, since the rock is dropped from rest off the bridge. Therefore, v_i=0 \frac{m}{s}.

We want to find the vertical displacement, the height of the bridge, so we can say that \triangle x= \ ?

We have 4 out of 5 variables:

  • v_i,\ a, \ t, \ \triangle x

Look through the constant acceleration equations to see which equation has all 4 of these variables. You should come up with this one (no final velocity):

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Subtract x_i from both sides of the equation to get:

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Substitute in our known variables and solve for delta x.

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0 m/s multiplied by 2.5 s is 0, so we have:

  • \triangle x =\frac{1}{2} (-10)(2.5)^2

Evaluate the exponent first and multiply the terms together.

  • \triangle x =(-5)(6.25)
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The vertical displacement is -31.25 meters from the rock's starting position, so we can say that the height of the bridge is 31.25 meters, which is approximately 31 meters tall.

7 0
3 years ago
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