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WARRIOR [948]
3 years ago
14

In a velocity selector having electric field E and magnetic field B, the velocity selected for positively charged particles is v

= E/B. The formula is the same for a negatively charged particles.
a. True
b. False
Physics
1 answer:
maw [93]3 years ago
6 0

Answer:

True or False

Explanation:

Because.....

easy 50% chance you are right

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If the PLATE SEPARATION of an isolated charged parallel-plate capacitor is doubled: A. the electric field is doubled
mash [69]

Explanation:

The electric field of an isolated charged parallel-plate capacitor is given by :

E=\dfrac{q}{A\epsilon_o}........(1)

Where

q is the electric charge

A is the area of cross section of parallel plate

It is clear from equation (1) that the electric field of a parallel plate capacitor is directly proportional to the charge on the plate and inversely proportional to the area of cross section of a plate.

So, the correct option is (E) i.e. "none of the above".

5 0
3 years ago
Faraday's Law states that the negative of the time rate of change of the flux of the magnetic field through a surface is equal t
MrRa [10]

Answer:

(C). The line integral of the magnetic field around a closed loop

Explanation:

Faraday's law states that induced emf is directly proportional to the time rate of change of magnetic flux.

This can be written mathematically as;

EMF = -\frac{\delta \phi _B}{\delta t}

(\frac{\delta \phi _B}{\delta t} ) is the rate of change of the magnetic flux through a surface bounded by the loop.

ΔФ = BA

where;

ΔФ is change in flux

B is the magnetic field

A is the area of the loop

Thus, according to Faraday's law of electric generators

∫BdL = \frac{\delta \phi _B}{\delta t} = EMF

Therefore, the line integral of the magnetic field around a closed loop is equal to the negative of the rate of change of the magnetic flux through the area enclosed by the loop.

The correct option is "C"

(C). The line integral of the magnetic field around a closed loop

8 0
4 years ago
If the Earth and distant stars were stationary (motionless) in space, what would we observe about the wavelength from these star
dangina [55]
1) In the first case, the correct answer is
<span>A.Wavelengths measured would match the actual wavelengths emitted.
In fact, the stars are not moving relative to Earth, so there is no shift in the measured wavelength.

2) In this second case, the correct answer is
</span><span>A.Wavelengths measured would be shorter than the actual wavelengths emitted.
</span>in fact, since the stars in this case are moving towards the Earth, then apparent frequency of their emitted light will be larger than the actual frequency, because of the Doppler effect, according to the formula:
f'= \frac{c}{c+v_s} f_0
where f0 is the actual frequency, f' the apparent frequency, c the speed of light and vs the velocity of the source (the stars) relative to the obsever (Earth). Vs is negative when the source is moving towards the observer, so the apparent frequency f' is larger than the actual frequency f0. But the wavelength is inversely proportional to the frequency, so the apparent wavelength will be shorter than the actual wavelength.
6 0
3 years ago
What is net force?
drek231 [11]
The answer is d my friend :)
5 0
3 years ago
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True or false, the hardest working muscle in the body is the gluteus Maximus?
Mademuasel [1]
False. I’m pretty sure the heart is the strongest muscle ( ._.)
8 0
3 years ago
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