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Vlad [161]
3 years ago
11

THE LARGEST HORIZONTAL PLATES IS SEPARATED BY 4mm .The plate is at the potencial of -6V . What potencial should be applied to th

e upper plate to create electric feild strength 4000Vm-1 ypward in the space between the plate
Physics
1 answer:
Serhud [2]3 years ago
6 0

Answer:

V₂ = -22 V

Explanation:

Electric potential and field are related

         ΔV = - E d

where ΔV is the potential difference between the plates, E the electric field and d the separation between the plates

 

In this exercise we are given the parcionero d = 4 mm = 0.004 m, the potential of one of the plates V1 = -6V and the value of the electric field E = 4000 V / m

          V₂- V₁ = - E d

          V₂ = - Ed + V₁

          V₂ = - 4000 0.004 + (- 6)

          V₂ = -16 - 6

          V₂ = -22 V

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Sound intensity :

The power carried by sound waves per unit area in the direction  perpendicular to that region is known as sound intensity or acoustic intensity. The watt per square meter (W/m2) is the SI unit of intensity, which also covers sound intensity. Sound intensity is a measure of how quickly energy moves across a given space. The unit area in the SI measurement system is 1 m2. So Watts per square meter are used to measure sound intensity. As there will be energy flow in certain directions but not in others, sound intensity also provides a measure of direction.

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1 year ago
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Answer:

50 meters

Explanation:

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Now that you have the acceleration, you can do this:

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Hope this helps!

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