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Andreyy89
3 years ago
9

Does current flow through or across a resistor?

Physics
2 answers:
Tanzania [10]3 years ago
6 0

It's not exactly clear what you think the difference is between "through" and "across".

A resistor has two wires.  Electric current that flows into one wire, continues <em>through</em> the entire body of the resistor and out <em>through</em> the other wire.  If there's a crack or break anywhere along the body of the resistor, the circuit will be 'open' and the current will stop flowing.

Now, if you were to connect a <em>voltmeter</em> between the ends of the resistor, the meter would measure and indicate the difference in electric potential between those two points.  That would be called the voltage '<u><em>across</em></u>' the resistor.  Numerically, it would be equal to the product of the resistor's resistance and the current through it.  

Mamont248 [21]3 years ago
3 0

Answer:

Current flows across a resistor.

Explanation:

Please mark brainliest and have a great day!

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Two coherent sources of radio waves, A and B, are 5.00 meters apart. Each source emits waves with wavelength 6.00 meters. Consid
Korolek [52]

Answer:

a

    z= 2.5 \ m

b

   z =  (1 \ m ,  4 \ m )

Explanation:

From the question we are told that

     Their distance apart is  d =  5.00 \ m

      The  wavelength of each source wave \lambda =  6.0 \ m

Let the distance from source A  where the construct interference occurred be z

Generally the path difference for constructive interference is

              z - (d-z) =  m \lambda

Now given that we are considering just the straight line (i.e  points along the line connecting the two sources ) then the order of the maxima m =  0

  so

        z - (5-z) =  0

=>     2 z - 5 =  0

=>     z= 2.5 \ m

Generally the path difference for destructive  interference is

           |z-(d-z)| = (2m + 1)\frac{\lambda}{2}

=>         |2z - d |= (0 + 1)\frac{\lambda}{2}

=>        |2z - d| =\frac{\lambda}{2}

substituting values

          |2z - 5| =\frac{6}{2}

=>      z  =  \frac{5 \pm 3}{2}

So  

      z =  \frac{5 + 3}{2}

      z =  4\ m

and

      z =  \frac{ 5 -3 }{2}

=>   z =  1 \ m

=>    z =  (1 \ m ,  4 \ m )

7 0
3 years ago
If
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Answer:

A factor of 2*4 = 8

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F_g = (G*m1*m2)/r^2  

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So, if you double m1 and quadruple m2:

m1' = 2*m1

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Then F_g' = (G*m1'*m2')/r^2 = (G*2*m1*4*m2)/r^2 = 8*(G*m1*m2)/r^2 = 8*F_g

3 0
2 years ago
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