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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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Explanation:

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. In order to prevent injury from inflating air bags, it is recommended that vehicle occupants sit at least __________ inches aw
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Small droplets of carbon tetrachloride at 68 °F are formed with a spray nozzle. If the average diameter of the droplets is 200 u
Licemer1 [7]

Answer:

the difference in pressure between the inside and outside of the droplets is 538 Pa

Explanation:

given data

temperature = 68 °F

average diameter = 200 µm

to find out

what is the difference in pressure between the inside and outside of the droplets

solution

we know here surface tension of carbon tetra chloride at 68 °F is get from table 1.6 physical properties of liquid that is

σ = 2.69 × 10^{-2} N/m

so average radius = \frac{diameter}{2} =  100 µm = 100 ×10^{-6} m

now here we know relation between pressure difference and surface tension

so we can derive difference pressure as

2π×σ×r = Δp×π×r²    .....................1

here r is radius and  Δp pressure difference and σ surface tension

Δp = \frac{2 \sigma }{r}    

put here value

Δp = \frac{2*2.69*10^{-2}}{100*10^{-6}}  

Δp = 538

so the difference in pressure between the inside and outside of the droplets is 538 Pa

7 0
3 years ago
A cylindrical insulation for a steam pipe has an inside radius rt = 6 cm, outside radius r0 = 8 cm, and a thermal conductivity k
goldfiish [28.3K]

Answer:

heat loss per 1-m length of this insulation is 4368.145 W

Explanation:

given data

inside radius r1 = 6 cm

outside radius r2 = 8 cm

thermal conductivity k = 0.5 W/m°C

inside temperature t1 = 430°C

outside temperature t2 = 30°C

to find out

Determine the heat loss per 1-m length of this insulation

solution

we know thermal resistance formula for cylinder that is express as

Rth = \frac{ln\frac{r2}{r1}}{2 \pi *k * L}   .................1

here r1 is inside radius and r2 is outside radius L is length and k is thermal conductivity

so

heat loss is change in temperature divide thermal resistance

Q = \frac{t1- t2}{\frac{ln\frac{r2}{r1}}{2 \pi *k * L}}

Q = \frac{(430-30)*(2 \pi * 0.5 * 1}{ln\frac{8}{6} }

Q = 4368.145 W

so heat loss per 1-m length of this insulation is 4368.145 W

4 0
3 years ago
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Lapatulllka [165]

Answer:

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A drum changes mechanical energy to <u>sound</u> energy

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A battery is a device that stores chemical energy and converts it to electrical energy.

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4 0
3 years ago
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