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sashaice [31]
3 years ago
12

Two points, A and B, are separated by 0.0158 m. The potential at A is +81.8 V, and that at B is +40.1 V. Find the magnitude of t

he constant electric field between the points.
Physics
1 answer:
Ostrovityanka [42]3 years ago
5 0

Answer:

2639.24 V/m

Explanation:

Electric field = potential difference between point A and B ÷ distance apart

Potential at A = 81.8 V

Potential at B = 40.1 V

Potential difference between A and B = 81.8 - 40.1 = 40.7 V

Distance apart = 0.0158 m

Electric field = 40.7 V ÷ 0.0158 m = 2639.24 V/m

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Explanation:

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KATRIN_1 [288]

Explanation:

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      \frac{W}{Q_{H_{1}}} = \frac{T_{H_{1}} - T_{C_{1}}}{T_{H_{1}}} ..... (1)

And, formula for maximum efficiency of Carnot refrigerator is as follows.

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Now, equating both equations (1) and (2) as follows.

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                    = \frac{T_{C_{2}}}{T_{H_{1}}} (\frac{T_{H_{1}} - T_{C_{1}}}{T_{H_{2}} - T_{C_{2}}})

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<span>1 Joule = 1 kg • (m/s)<span>2
</span></span>the force as a function of mass of the object.
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I'm willing to give a lot of points if u can help me out.
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Answer:

i have absolutly no idea how to do it but i looked it up and your answer should be B. i could be wrong but thats what the web told me

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