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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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Why does the auto industry prefer uniform (national) standards for automobile emissions as opposed to regionally varying standar
WINSTONCH [101]

Answer:

Explanation:

For automobile emission, a uniform standard is preferred, because no unnecessary advantage is given by it to any company that is located in particular states where the regional standards are less severe.

Since pollution has its impact across the states and in the whole of the USA, then there should be uniform standards across all the states. It will also invalidate the impact of regional standards as a factor in the selection of plant locations for the automobile company. It means that a state offering less valid emission standards, will attract more companies to herself and it will be against the other states who care more about the natural environment. It can make more states to opt for the permissive emission standards, that will be more harmful to the USA as a country, than the good. So, a uniform standard is preferred to eliminate it as a factor in plant location decisions.

Yes, uniform standards are beneficial to everyone, because it will bring effective control upon the pollution level because there will be no state where the culprit firm can hide. Besides, it is more effective as efforts done towards environment conservation.

3 0
3 years ago
Whats the difference between GeForce GTX 1060 and Geforce GTX 3060? Is there any big changes to FPS and other settings?
ZanzabumX [31]

Answer:

uhhhhh, are you kidding? a GTX 3060 is far better than a 1060 ding dong

Explanation:

7 0
3 years ago
Read 2 more answers
A labor-intensive process to manufacture a product has a fixed cost of $338,000 and a variable cost of $143 per unit. An automat
ozzi

Answer:

no of unit is 17941

Explanation:

given data

fixed cost = $338,000

variable cost = $143 per unit

fixed cost = $1,244,000  

variable cost = $92.50 per unit

solution

we consider here no of unit is = n

so here total cost of labor will be sum of fix and variable cost i.e

total cost of labor = $33800 + $143 n  ..........1

and

total cost of capital intensive  = $1,244,000 + $92.5 n   ..........2

so here in both we prefer cost of capital if cost of capital intensive less than cost of labor

$1,244,000 + $92.5 n  <  $33800 + $143 n

solve we get

n > \frac{906000}{50.5}

n > 17941

and

cost of producing less than selling cost so here

$1,244,000 + $92.5 n < 197 n

solve it we get

n > \frac{1244000}{104.5}  

n > 11904

so in both we get greatest no is 17941

so no of unit is 17941

3 0
3 years ago
A gas turbine operates with a regenerator and two stages of reheating and intercooling. Air enters this engine at 14 psia and 60
Rzqust [24]

Answer:

flow(m) = 7.941 lbm/s

Q_in = 90.5184 Btu/lbm

Q_out = 56.01856 Btu/lbm

Explanation:

Given:

- T_1 = 60 F = 520 R

- T_6 = 940 = 1400 R

- Heat ratio for air k = 1.4

- Compression ratio r = 3

- W_net,out = 1000 hp

Find:

mass flow rate of the air

rates of heat addition and rejection

Solution:

- Using ideal gas relation compute T_2, T_4, T_10:

                     T_2 = T_1 * r^(k-1/k)

                     T_2 = T_4 = T_10 = 520*3^(.4/1.4) = 711.744 R

- Using ideal gas relation compute T_7, T_5, T_9:

                     T_7 = T_6 * r^(-k-1/k)

                     T_7 = T_5 = T_9 = 1400*3^(-.4/1.4) = 1022.84 R

- The mass flow rate is obtained by:

                     flow(m) = W_net,out / 2*c_p*(1400-1022.84-711.744+520)

                     flow(m) = 1000*.7068 / 2*0.24*(1400-1022.84-711.744+520)

                     flow(m) = 7.941 lbm/s

- The heat input is as follows:

                     Q_in = c_p*(T_6 - T_5)

                     Q_in = 0.24*(1400 - 1022.84)

                     Q_in = 90.5184 Btu/lbm

- The heat output is as follows:

                     Q_out = c_p*(T_10 - T_1)

                     Q_out = 0.24*(711.744 - 520)

                    Q_out = 56.01856 Btu/lbm

                                           

                     

5 0
3 years ago
Cuál de las siguientes es la mejor manera de practicar sus habilidades de tecnología de secundaria?
Mkey [24]
Huh? Do you know English?
8 0
3 years ago
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