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jolli1 [7]
3 years ago
14

A 330 μf capacitor is connected in series with a 120 ohm resistor. what is the time constant for the circuit?

Physics
1 answer:
n200080 [17]3 years ago
7 0
Time constant of a RC circuit is

Resistance x Capacitance = 120 x [330 x 10^(-6)] = 0.0396 s
You might be interested in
Explain why nuclear fission and nuclear fusion release large amounts of energy
levacccp [35]

Answer:

Because of the formula E=mc^2

Explanation:

In this problem we are describing two different processes:

  • Nuclear fission occurs when a heavy, unstable nucleus breaks apart into two or more lighter nuclei
  • Nuclear fusion occurs when two (or more) light nuclei fuse together producing a heavier nucleus

In both cases, the total mass of the final products is smaller than the total mass of the initial nuclei.

According to Einsten's formula, this mass difference has been converted into energy, as follows:

E=\Delta mc^2

where:

E is the energy released in the reaction

\Delta m is the mass defect, the difference between the final total mass and the initial total mass

c=3.0 \cdot 10^8 m/s is the speed of light

From the formula, we see that the factor c^2 is a very large number, therefore even if the mass defect \Delta m is very small, nuclear fusion and nuclear fission release huge amounts of energy.

8 0
3 years ago
How do the dark lines of an atom''s absorption spectrum relate to the bright lines of its emission spectrum?
tangare [24]

Wouldn't it be neat if an electron falling closer to the nucleus ... emitting a
photon ... actually gave out more energy than it needed to climb to its original
energy level by absorbing a photon !   If there were some miraculous substance
that could do that, we'd have it made.

All we'd need is a pile of it in our basement, with a bright light bulb over the pile,
connected to a tiny hand-crank generator.

Whenever we wanted some energy, like for cooking or heating the house, we'd
switch the light bulb on, point it towards the pile, and give the little generator a
little shove.  It wouldn't take much to git 'er going.

The atoms in the pile would absorb some photons, raising their electrons to higher
energy levels.  Then the electrons would fall back down to lower energy levels,
releasing more energy than they needed to climb up.  We could take that energy,
use some of it to keep the light bulb shining on the pile, and use the extra to heat
the house or run the dishwasher.

The energy an electron absorbs when it climbs to a higher energy level (forming
the atom's absorption spectrum) is precisely identical to the energy it emits when
it falls back to its original level (creating the atom's emission spectrum).

Energy that wasn't either there in the atom to begin with or else pumped
into it from somewhere can't be created there.

You get what you pay for, or, as my grandfather used to say, "For nothing
you get nothing."

3 0
3 years ago
You are designing an optical fiber scope for directing light into a confined area. You want to keep light within the fiber. Base
shepuryov [24]

Answer:

Explanation:

 For entry of light into tube of unknown refractive index

sin ( 90 - 25 ) / sinr = μ , μ is the refractive index of the tube , r is angle of refraction in the medium of tube

r = 90 - C where C is critical angle between μ and body medium in which tube will be inserted.

sin ( 90 - 25 ) / sin( 90 - C)  = μ

sin65 / cos C = μ

sinC = 1.33 / μ  , where 1.33 is the refractive index of body liquid.

From these equations

sin65 / cos C = 1.33 / sinC

TanC = 1.33 / sin65

TanC = 1.33 / .9063

TanC = 1.4675

C= 56°

sinC = 1.33 / μ

μ = 1.33 / sinC

= 1.33 / sin56

= 1.33 / .829

μ = 1.6   Ans

3 0
3 years ago
If a galaxy has an apparent velocity of 2300 km/s, what is its distance if the Hubble constant is assumed to be 70 km/s/Mpc
prohojiy [21]

The distance of the galaxy is 32.86 Mpc.

Using the hubble law, v = H₀D where v = apparent velocity of galaxy = 2300 km/s, H = hubble constant = 70 km/s/Mpc and D = distance of galaxy.

Since we require the distance of the galaxy, we make D subject of the formula in the equation. So, we have

D = v/H₀

Substituting the values of the variables into the equation, we have

D = 2300 km/s ÷ 70 km/s/Mpc

D = 32.86 Mpc

So, the distance of the galaxy is 32.86 Mpc

Learn more about hubble law here:

brainly.com/question/18484687

4 0
2 years ago
HELP ME PLSSS I WILL GIVE YOU ANYTHING HELPPPPPPPP MEEEEEEEE
Minchanka [31]

Answer:

B)

Explanation:

I hope this help's :)

5 0
3 years ago
Read 2 more answers
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