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tatuchka [14]
3 years ago
9

A generator contains a 100-turn coil that rotates 10.0 times per second. if each turn has an area of 0.100 m2 and the magnetic f

ield through the coils is 0.050 t, what is the maximum emf induced in the coil?
Physics
1 answer:
Andrews [41]3 years ago
6 0
We use Lenz's law:
emf = -(NBA)/t ; N is the number of turns, B is the magnetic field, A is the area
t = 1/f = 1/10
emf = (-100 x 0.05 x 0.1) / 0.1
emf = -5 V
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4. The atmosphere is composed of about 78% nitrogen, 21% oxygen, and 1% argon. Typical atmospheric pressure in Boulder, Colorado
photoshop1234 [79]

<u>ANS</u>

<u>Step</u> <u>1</u> :Explanation of required formula.

According to Dalton's Law of Partial Pressures, the partial pressure of a component of a gaseous mixture depends on the mole ratio of said component and the total pressure of the gaseous mixture

i.e Pi=Xi × Ptotal,

Here we don't know exactly how many moles of the mixture we have,

but we know that 78.0% of all the molecules present in the mixture are nitrogen molecules, 21.0% are oxygen molecules, and 1% are molecules of Ar gas.

As we know, a mole is simply a very large collection of molecules. In order to have one mole of a substance, we need to have 6.022 × 1023 molecules of that substance.

This means that the actual number of moles is not important here, because the ratio that exists between the number of molecules is equivalent to the ratio that exists between the number of moles.

Hence,

<u>Step</u> <u>2</u> : Calculate of mole fraction of the mixture.

mole fraction of nitrogen = 78 /100 = 0.78

mole fraction of O2 =

21 /100 = 0.21

mole fraction of Argon =

1 /100 = 0.01.

<u>Step</u> <u>3</u> : Calculate the pressure contributed by each of the mixture.

The pressure contributed by N2 = mole fraction of N2 × Total pressure = 0.78 × 0.83 atm = 0.6474 atm

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The pressure contributed by N2 = 0.01 × 0.83 atm = 0.0083 atm.

<u>Tha</u><u>nk</u> <u>You</u> !!!!!!

6 0
2 years ago
Which of these words describes humans
MA_775_DIABLO [31]
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3 0
3 years ago
A particle with charge 3.01 µC on the negative x axis and a second particle with charge 6.02 µC on the positive x axis are each
ra1l [238]

Answer:

The third particle should be at 0.0743 m from the origin on the negative x-axis.

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We know that the electric field is:

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So, we have:

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Let's solve it for r(3).

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r_{3}=0.0743\:  

Therefore, the third particle should be at 0.0743 m from the origin on the negative x-axis.

I hope it helps you!

 

3 0
2 years ago
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