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Anna [14]
2 years ago
13

In a certain region of space, the electric field is zero. from this fact, what can you conclude about the electric potential in

this region?
Physics
1 answer:
eduard2 years ago
5 0
The answer is that it is constant. The relation between electric field and electric potential is given as, E=  -gradient (V).  The E, the partial rate of change of Electric potential, in the equation implies that the V, the partial differential of the potential of the three-dimensional space (assuming it is considered) is constant. 
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zloy xaker [14]

Answer:

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Explanation:

6 0
2 years ago
Which list places the layers of the sun in the correct order from outermost to innermost?
Readme [11.4K]

Answer:

The Sun's layers consist of the following in this order.

1) Corona

2) Transition Region

3) Chromosphere

4) Photosphere

5) Convection Zone

6) Radiative Zone

and last but not least 7) The Core

Hope this helps ;)

6 0
3 years ago
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Degger [83]

Answer:

why would you waste points

Explanation:

4 0
2 years ago
Read 2 more answers
If you push a crate across a factory floor at constant speed in a constant direction, what is the magnitude of the force of fric
poizon [28]

Answer:

The magnitude of the force of friction equals the magnitude of my push

Explanation:

Since the crate moves at a constant speed, there is no net acceleration and thus, my push is balanced by the frictional force on the crate. So, the magnitude of the force of friction equals the magnitude of my push.

Let F = push and f = frictional force and f' = net force

F - f = f' since the crate moves at constant speed, acceleration is zero and thus f' = ma = m (0) = 0

So, F - f = 0

Thus, F = f

So, the magnitude of the force of friction equals the magnitude of my push.

3 0
2 years ago
A concave lens has a focal length of 25cm. it's power in diaptor is​
IgorLugansk [536]

As we know that :

\begin{gathered}\large{\boxed{\sf{P \: = \: \dfrac{1}{f}}}} \\ \\ \rightarrow {\sf{P \: = \: \dfrac{1}{-25}}}\end{gathered}

Power, is in Meter. So divide focal length by 100

\begin{gathered}\rightarrow {\sf{P \: = \: \dfrac{1}{\dfrac{-25}{100}}}} \\ \\ \rightarrow {\sf{P \: = \: \dfrac{-100}{25}}} \\ \\ \rightarrow {\sf{P \: = \:- 4}} \\ \\ \underline{\sf{\therefore \: Power \: of \: Concave \: lens \: is \: - \: 4D}}\end{gathered}

8 0
1 year ago
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