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3 years ago

A constant magnetic flux through a closed loop of wire induces an emf in that loop. True or false?

Physics

1 answer:

miskamm [114]3 years ago

6 0

Answer: False

Explanation: In order to explain this problem we have to use the Faraday law, which say

dФm/dt=-ε it means that the variation of the magnetic field flux with time is equal to the emf ( electromotive force). In our case the magnetic flux is constant then there is not a emf induced in a wire closed loop.

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An ocean wave has an amplitude of 2.5 m. Weather conditions suddenly change such that the wave has an

Dovator [93]

Answer:

2.5

Explanation:2.5 +2.5 = 5.0

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3 years ago

At what time of year is the intensity of solar radiation striking each of earth's hemispheres weakest?

bagirrra123 [75]

<span>During winter for a given hemisphere, solar radiation reaches the lowest period of its annual cycle due to the tilt of the earth on its axis. As the earth rotates around the sun, this tilt occludes a portion of the energy released by the sun as it diffuses in the atmosphere.</span>

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3 years ago

If an engine does 660 J of work in 10 seconds, its average power is ...

givi [52]

Answer:

66w

Explanation:

p=w/t

p=660/10

p=66

prolly a bad explanation but hope it helps...

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2 years ago

Hi!

dangina [55]

Answer:

the moe weight you have in the marble, the higher the speed on the way down

Explanation:

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3 years ago

One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories

ch4aika [34]

Answer:

114.92749 keV

Explanation:

r = Radius of trajectory

m = Mass of electron = $9.11\times 10^{-31}\ kg$

B = Magnetic field = 0.044 T

q = Charge of electron = $1.6\times 10^{-19}\ C$

The centripetal force and the magnetic forces are conserved

$m\frac{v^2}{r}=Bqv\\\Rightarrow v=\frac{Bqr}{m}$

Velocity of first electron

$v=\frac{Bqr_1}{m}\\\Rightarrow v=\frac{0.044\times 1.6\times 10^{-19}\times 0.01}{9.11\times 10^{-31}}\\\Rightarrow v_1=77277716.79473\ m/s$

Velocity of second electron

$v=\frac{Bqr_2}{m}\\\Rightarrow v_2=\frac{0.044\times 1.6\times 10^{-19}\times 0.024}{9.11\times 10^{-31}}\\\Rightarrow v_2=185466520.30735\ m/s$

Total kinetic energy is given by

$K=K_1+K_2\\\Rightarrow K=\frac{1}{2}mv_1^2+\frac{1}{2}mv_2^2\\\Rightarrow K=\frac{1}{2}m(v_1^2+v_2^2)\\\Rightarrow K=\frac{1}{2}\times 9.11\times 10^{-31}(77277716.79473^2+185466520.30735^2)\\\Rightarrow K=1.83884\times 10^{-14}\ J$

Converting to eV

$1\ J=\frac{1}{1.6\times 10^{-19}}\ eV$

$1.83884\times 10^{-14}\ J=1.83884\times 10^{-14}\times \frac{1}{1.6\times 10^{-19}}\ eV\\ =114927.49\ ev=114.92749\ keV$

The energy of incident electron is 114.92749 keV

5 0

3 years ago