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Vlad [161]
3 years ago
10

A constant magnetic field passes through a single rectangular loop whose dimensions are 0.41 m x 0.65 m. The magnetic field has

a magnitude of 1.8 T and is inclined at an angle of 78° with respect to the normal to the plane of the loop.
(a) If the magnetic field decreases to zero in a time of 0.63 s, what is the magnitude of the average emf induced in the loop?
(b) If the magnetic field remains constant at its initail calue of 2.1T, what is the magnitude of the rate ΔA/Δt at which the area should change so that the average emf has the same magnitude as in part (a)?
Physics
1 answer:
Kruka [31]3 years ago
6 0

Answer:

hfghgfhg

Explanation:

ghfhghfhhfgj

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Answer:

825 m

Explanation:

330 m/s

1 / 0.40 = 2.5

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What does friction oppose
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Explanation:

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To meet a U.S. Postal Service requirement, employees' footwear must have a coefficient of static friction of 0.5 or more on a sp
liq [111]

Answer:

Minimum time interval (t2)=0.90 SECONDS

Explanation:

  • coefficient of friction for employees footwear = 0.5
  • coefficient of friction for typical athletic shoe = 0.810
  • frictional force = coefficient of friction X acceleration due to gravity X mass of body
  • Acceleration due to gravity is a constant = 9.81 m/s
  • Let frictional force for employee footwear = FF1
  • Let frictional force for athletic footwear =FF2

                 FF1 = O.5 X 9.81 X mass of body

                         = 4.905 x mass of body

                  FF2 = 0.810 X 9.81 X mass of body

                          = 7.9461 x mass of body

The body started from rest there by making the initial velocity zero ( u = 0)

From d= ut + 1/2 a x t^{2}

  •      d = \frac{1}{2} x a x t^{2}  .....................................i  

            where d= distance and it is given as 3.25m

  •          F =ma  ...................................ii

making acceleration subject of the formula from equation ii

  •              a =\frac{F}{m}

         Making t subject of formula from equation (i)

  • t=\sqrt{\frac{2d}{(f/m} }

    where

  • \frac{FF1}{Mass of body} = 4.905
  • \frac{FF2}{Mass of body} =7.9461

  Let

  •            t1 = minimum time taken for frictional force for employee foot wear
  •                                 t1 = \sqrt{\frac{6.5}{4.905} } =1.15 seconds

  •                                  t2 = \sqrt{\frac{6.5}{7.9461} } = 0.90 seconds

 

THANK YOU

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