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Ber [7]
3 years ago
15

How much work does the electric field do in moving a proton from a point with a potential of +V1 = +185 V to a point where it is

V2 = 55.0 V?
Engineering
1 answer:
Arada [10]3 years ago
4 0

Answer:

W=  2.08 x 10⁻¹⁷  J

Explanation:

Given that

Initial potential V₁ =185 V

Final potential V₂ = 55 V

We know that charge of the proton

q=1.6 x 10⁻¹⁹ C

Work done is given as

W= q ΔV

q=Charge

ΔV=Potential difference

W=Work done

Now by putting the values in the above equation then we get

W= 1.6 x 10⁻¹⁹  ( 185 - 55 ) J

W=208 x 10⁻¹⁹ J

W=  2.08 x 10⁻¹⁷  J

Therefore the work done will be 2.08 x 10⁻¹⁷  J.

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Answer:

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You have been assigned the task of reviewing the relief scenarios for a specific chemical reactor in your plant. You are current
postnew [5]

Answer:

D=0.016m

Explanation:

From the question we are told that:

Discharge Rate F_r=0.5kgls

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Generally the equation for Density is mathematically given by

\rho=\frac{PM}{RT}

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Generally the equation for Flow rate is mathematically given by

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Q=Heat coefficient\ ratio\ of\ Nitrogen

Q=1.4

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Therefore

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alina1380 [7]

Answer:

The engineering design process

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