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Wittaler [7]
3 years ago
7

A coil of wire with 60 turns lies on the plane of the page, and a uniform magnetic field points into the page with a strength of

1.10 T. The coil has an initial area of 0.890 m2 and is then elongated so that its area decreases to 0.270 m2 in 0.80 s. What is the magnitude of the average induced emf in the coil while its area is changing? 59.334 Incorrect: Your answer is incorrect. V
Physics
1 answer:
baherus [9]3 years ago
4 0

Answer:

0.85V

Explanation:

The emf is calculated by using the Lenz's Law

\epsilon=-\frac{\Delta\Phi_B}{\Delta t}

But for this case we have that the magnetic field is constant. Hence we have

\epsilon=-\frac{B(A_{2}-A_{1})}{t_2-t_1}=-\frac{(1.10T)(0.270m^{2}-0.89m^{2})}{0.8s}\\\\\epsilon=0.85V

where we have taken that the intial time is t1=0

I hope this is useful for you

regards

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The car will gain new momentum if it's velocity is doubled or tripled.
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10. The electron dot diagram for the element Ne would have ​
zmey [24]

Answer:

2,8

Explanation:

The first electron shell would have 2 electrons, the second shell would have 8 electrons. This is because Neon has a relative charge of 10.

7 0
3 years ago
Drew wants to convert square mils to circular mils in an area calculation. How does he do this?
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Answer:

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The viewing screen in a double-slit experiment with monochromatic light. Fringe C is the central maximum. The fringe separation
makvit [3.9K]

Answer:

<em>Part A</em><em>:</em>

a) If the wavelength of the light is decreased the fringe spacing Δy will decrease.

<em>Part B</em><em>:</em>

b) If the spacing between the slits is decreased the fringe spacing Δy will increase.

<em>Part C</em><em>:</em>

a) If the distance to the screen is decreased the fringe spacing will decrease.

<em>Part D</em><em>:</em>

The dot in the center of fringe E is 920\ x\ 10^{-9} m farther from the left slit than from the right slit.

Explanation:

In the double-slit experiment there is a clear contrast between the dark and bright fringes, that indicate destructive and constructive interference respectively, in the central peak and then is less so at either side.

The position of bright fringes in the screen where the pattern is formed can be calculated with

                      \vartriangle y =\frac{m \lambda L}{d}

                      m=0,\pm 1,\pm 2,\pm 3,.....

  1. m is the order number.
  2. \lambda is the wavelength of the monochromatic light.
  3. L is the distance between the screen and the two slits.
  4. d is the distance between the slits.
  • Part A:  a) In the above equation for the position of bright fringes we can see that if the wavelength of the light \lambda is decreased the overall effect will be that the fringes are going to be closer. That means that the fringe spacing Δy will decrease.
  • Part B:  b) In the above equation for the position of bright fringes we can see that if the spacing between the slits d is decreased the fringes are going to be wider apart. That means the fringe spacing Δy will increase.
  • Part C:  a) In the above equation we can see that if the distance to the screen L is decreased the fringes are going to be closer. That means the fringe spacing Δy will decrease.
  • Part D: We are told that the central maximum is the fringe C that corresponds with m=0. That means that fringe E corresponds with the order number m=2 if we consider it to be the second maximum at the rigth of the central one. To calculate how much farther from the left slit than from the right slit is a dot located at  the center of the fringe E in the screen we use the condition for constructive interference. That says that the  path length difference Δr between rays coming from the left and right slit must be \vartriangle r=m \lambda

        We simply replace the values in that equation :

                      \vartriangle r= m \lambda =2.\ 460\ nm

                      \vartriangle r= 920\ x\ 10^{-9} m

         The dot in the center of fringe E is 920\ x\ 10^{-9}m farther from the left slit than from the right slit.

     

       

       

     

3 0
3 years ago
1) Arrange the following spectral regions in order of increasing energy: infrared, microwave, ultraviolet, visible.
Anton [14]

Answer:

The order of increasing energy is as follows

"microwave < infrared < visible < ultraviolet"

Option (A) is correct.

Explanation:

Given:

Arrange the following spectral regions in order of increasing energy: infrared, microwave, ultraviolet, visible.

From the formula of energy in terms of frequency.

  E = hf

Where h = planck constant, f = frequency of light.

From above formula we can conclude that higher frequency means higher energy.

In our case ultraviolet has higher frequency and microwave has lower frequency.

So ultraviolet has higher energy and microwave has lower energy.

microwave < infrared < visible < ultraviolet

Therefore, the order of increasing energy is as follows

"microwave < infrared < visible < ultraviolet"

7 0
3 years ago
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