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velikii [3]
2 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]2 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]2 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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un tanque de gasolina de 40 litros fue llenado por la noche, cuando la temperatura era de 68 grados farenheit. Al dia siguiente
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Volume of gasoline that expands and spills out is 1.33 ltr

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here we know that

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3 0
2 years ago
A baseball thrown at an angle of 60.0° above the horizontal strikes a building 16.0 m away at a point 8.00 m above the point fro
yanalaym [24]

Answer:

a) v_{o} =16m/s

b) v=9.8m/s

c) \beta =-35.46º

Explanation:

From the exercise we know that the ball strikes the building 16m away and its final height is 8m more than the initial

Being said that, we can calculate the initial velocity of the ball

a) First we analyze its horizontal motion

x=v_{ox}t

x=v_{o}cos(60)t

v_{o}=\frac{x}{tcos(60)}=\frac{16m}{tcos(60)} (1)

That would be our first equation

Now, we need to analyze its vertical motion

y=y_{o}+v_{oy}t+\frac{1}{2}gt^2

y_{o}+8=y_{o}+v_{o}sin(60)t-\frac{1}{2}(9.8)t^2

Knowing v_{o} in our first equation (1)

8=\frac{16}{tcos(60)}sin(60)t-\frac{1}{2}(9.8)t^2

\frac{1}{2}(9.8)t^2=16tan(60)-8

Solving for t

t=\sqrt{\frac{2(16tan(60)-8)}{9.8} } =2s

So, the ball takes to seconds to get to the other building. Now we can calculate its <u>initial velocity</u>

v_{o}=\frac{16m}{(2s)cos(60)}=16m/s

b) To find the <u>magnitude of the ball just before it strikes the building</u> we need to calculate its x and y components

v_{x}=v_{ox}+at=16cos(60)=8m/s

v_{y}=v_{oy}+gt=16sin(60)-(9.8)(2)=-5.7m/s

So, the magnitude of the velocity is:

v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{(8m/s)^2+(-5.7m/s)^2}=9.8m/s

c) The <u><em>direction of the ball</em></u> is:

\beta=tan^{-1}(\frac{v_{y} }{v_{x}})=tan^{-1}(\frac{-5.7}{8})=-35.46º

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Nana76 [90]

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