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Bess [88]
3 years ago
8

A magnetic field is uniform over a flat, horizontal circular region with a radius of 1.50 mm, and the field varies with time. In

itially the field is zero and then changes to 1.50 T, pointing upward when viewed from above, perpendicular to the circular plane, in a time of 130 ms. What is the average induced emf around the border of the circular region?
Physics
1 answer:
Daniel [21]3 years ago
3 0

Answer:81.57\mu V

Explanation:

Given

radius of circular region r=1.50 mm

A=\pi r^2=7.069\times 10^{-6}\ m^2

Magnetic Field B=1.50\ T

time t=130 ms

Flux is given by

\phi =B\cdot A

change in Flux d\phi =(B_f-B_i)A

Emf induced is e=\frac{\mathrm{d} \phi}{\mathrm{d} t}

e=\frac{(1.5)\cdot 7.069\times 10^{-6}}{130\times 10^{-3}}

e=81.57 \mu V

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What effect do sound waves have on air particles
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Answer:

Sound waves travel at 343 m/s through the air and faster through liquids and solids. The waves transfer energy from the source of the sound, e.g. a drum, to its surroundings. Your ear detects sound waves when vibrating air particles cause your ear drum to vibrate. The bigger the vibrations the louder the sound.

Explanation:

7 0
3 years ago
Two homogeneous bodies of the same volume
lord [1]

Answer:

No, it is not necessary for them to have same mass.

Explanation:

Let both bodies have a density d1 and d2 respectively.

Since their volumes are equal V1 = V2

we know that, https://tex.z-dn.net/?f=%5Cfrac%7Bmass%7D%7Bvolume%7D

Hence, d1 =  and d2 =    

Taking the ratio of densities,we get

This implies that unless the bodies have same densities, the mass of the two bodies will not be same.

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2 years ago
10. A frog leaps from its resting position at the
lana [24]

Answer:

The answer is C

Explanation:

F = m × a

m = 0.5kg

a = 3m/s²

F = 0.5 × 3

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5 0
3 years ago
Consider an isolated system, by which we mean a system free of external forces. If the sum of the particle energies in the syste
Airida [17]

Answer:

The system's potential energy is -147 J.

Explanation:

Given that,

Energy = 147 J

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System is isolated and it is free from external forces.

So, the work done by the external forces on the system should be equal to zero.

W=0

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Using thermodynamics first equation

\Delta U=W-\Delta E

Put the value into the formula

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Hence, The system's potential energy is -147 J.

5 0
2 years ago
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
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