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Ksenya-84 [330]
3 years ago
10

Please help!!!!!!!!!!!!!!!!!!!!!!

Physics
2 answers:
Doss [256]3 years ago
8 0

Answer:

Precipitate

Explanation:

KonstantinChe [14]3 years ago
7 0

Answer: I'm sure the answer is a.

Explanation: I hope that helped you.

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photoshop1234 [79]

I think it is Drift Velocity.

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jarptica [38.1K]

Answer:

5.03

Explanation:

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How do you calculate the fundamental frequency of a pipe?
ankoles [38]

This depends on whether the pipe is closed or open ended.

The fundamental frequency of a pipe is the simplest, smallest portion of a wave that can fit into a pipe. At the open end of a pipe, there is always an antinode - an area with maximum air movement.

If it is an open ended pipe, there is an antinode at each end, meaning that the length of the pipe is equal to 1/2 <span>λ </span>. Manipulating the formula <span><span>v=fλ</span> </span> to solve for the fundamental frequency leaves us with <span><span>f=<span>v/<span>2L</span></span></span> </span> in an open ended pipe.

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Please help!
Yanka [14]

Answer:

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7 0
3 years ago
The magnetic flux through a metal ring varies with time t according to ΦB = at3 − bt2, where ΦB is in webers, a = 6.00 Wb s−3, b
pychu [463]

To solve this problem it is necessary to apply the second derivative of the function to find the maximum time reference and thus calculate the maximum voltage.

With the maximum voltage by Ohm's Law it is possible to find the maximum current.

Ohm's law defines that

E = I*R

Where,

I = Current

R= Resistance

On the other hand by faraday studies and the potential can be expressed at the rate of change of the electric flow, that is

E = \frac{d\phi}{dt}

Replacing with our values we have that

E = \frac{d(6t^3-18t^2)}{dt}

E = -18t^2 +36t

The second derivative is

E' = -36t+36

When E' = 0 we have a Maximum, then

0 = -36t+36

t = 1

Therefore when the time is 1s E has a Maximum, replacing at the function

E(t) = -18t^2 +36t

E(1) = -18(1)^2 +36(1)

E = 18V

Then the maximum current will be given by

I = \frac{E}{R}

I = \frac{18}{2.8}

I = 6.42A

Therefore the maximum current induced in the ring is 6.42A

6 0
4 years ago
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