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wariber [46]
3 years ago
13

How much work is required to move an electron from the positive terminal to the negative terminal of a 12 V battery? (Knight 21.

4) A 20 nC charge is moved from a point where V=150V to point where V=-50V. How much work is done to move the charge?
Engineering
1 answer:
Ray Of Light [21]3 years ago
6 0

Answer:

work which is required to move an electron from the positive terminal to the negative terminal is

-4μJ

W = -4μJ

Explanation:

Work required to move the charge

W = qΔV

initial point V₁ = 150V

Final point V₂ = -50V

W = q(V₂ -V₁)

= 20 × 10⁹(-50 - 150)

W = -4μJ

work which is required to move an electron from the positive terminal to the negative terminal is

-4μJ

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A refrigerated space is maintained at -15℃, and cooling water is available at 30℃, the refrigerant is ammonia. The refrigeration
Illusion [34]

Answer:

(1) 5.74

(2) 5.09

(3) 3.05×10⁻⁵ kg/s

(4) 0.00573 kW

Explanation:

The parameters given are;

Working temperature, T_C  = -15°C = 258.15 K

Temperature of the cooling water, T_H = 30°C = 303.15 K

(1) The Carnot coefficient of performance is given as follows;

\gamma_{Max} = \dfrac{T_C}{T_H - T_C}  =  \dfrac{258.15}{303.15 - 258.15}   = 5.74

(2) For ammonia refrigerant, we have;

h_2 = h_g = 1466.3 \ kJ/kg

h_3 = h_f = 322.42 \ kJ/kg

h_4 = h_3 = h_f = 322.42 \ kJ/kg

s₂ = s₁ = 4.9738 kJ/(kg·K)

0.4538 + x₁ × (5.5397 - 0.4538) = 4.9738

∴ x₁ = (4.9738 - 0.4538)/(5.5397 - 0.4538) = 0.89

h_1 = h_{f1} + x_1 \times h_{gf}

h₁ = 111.66 + 0.89 × (1424.6 - 111.66) = 1278.5 kJ/kg

\gamma = \dfrac{h_1 - h_4}{h_2 - h_1}

\gamma = \dfrac{1278.5 - 322.42}{1466.3 - 1278.5} = 5.09

(3) The circulation rate is given by the mass flow rate, \dot m as follows

\dot m = \dfrac{Refrigeration \ capacity}{Refrigeration \ effect \ per \ unit \ mass}

The refrigeration capacity = 105 kJ/h

The refrigeration effect, Q = (h₁ - h₄) = (1278.5 - 322.42) = 956.08 kJ/kg

Therefore;

\dot m = \dfrac{105}{956.08}  = 0.1098 \ kg/h

\dot m = 0.1098 kg/h = 0.1098/(60*60) = 3.05×10⁻⁵ kg/s

(4) The work done, W = (h₂ - h₁) = (1466.3 - 1278.5) = 187.8 kJ/kg

The rating power = Work done per second = W×\dot m

∴ The rating power = 187.8 × 3.05×10⁻⁵ = 0.00573 kW.

6 0
3 years ago
To solve the problem, make assumptions for missing data and justify. Given:
finlep [7]

Answer:

5,4,1, this is a explication

6 0
3 years ago
A large retail organization previously divided work among its four employee benefits staff into distinct specializations. One pe
denpristay [2]

Answer:

This job restructuring is an example of job enrichment.

Explanation:

Job enrichment is a motivational tool used by organizations to give employees greater satisfaction in their jobs, by giving them additional responsibilities previously reserved for higher-ranking positions.

8 0
3 years ago
Which do you think would cause the most destruction to the organisms in a food web: taking away the carnivores, taking away the
kotegsom [21]

Answer:

Explanation:

I do not have access to your diagram but based on what I know, a food web cannot sustain itself without decomposers. They are the main guys who break down dead animals, plants into organic or inorganic nutrients which are needed by the primary producers to grow. So if decomposers are not there then producers cannot produce and thus herbivores cannot live and carnovers die out as well. Hope this makes sense.

Check out: https://www.nationalgeographic.org/encyclopedia/decomposers/

6 0
3 years ago
3. If nothing can ever be at absolute zero, why does the concept exist?
JulijaS [17]
The reason has to do with the amount of work necessary to remove heat from a substance, which increases substantially the colder you try to go. To reach zero kelvins, you would require an infinite amount of work.
6 0
2 years ago
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