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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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Determine the complex power, apparent power, average power absorbed, reactive power, and power factor (including whether it is l
LenKa [72]

Answer:

A) complex power = apparent  power =  ( 108.253 + 62.5 i ) VA  

    active power = 108.25 watts

   reactive power = 62.5 VAR'S

    power factor = cos ∅ = cos 30° = 0.866 ( lagging )

also ; current lags voltage by 30°

B) your question is not well written hence no answer

Explanation:

A) v(t) = 100 cos (377t - 30° ) v

    Vrms = \frac{100}{\sqrt{2} } ∠ -30°

    I(t) = 2.5cos(377t- 60°) A

  Irms = \frac{2.5}{\sqrt{2} } ∠ -60°

determine complex power apparent power . ......  power factor

Note : complex power = apparent power

=( Irms ) * Vrms

=  ( \frac{2.5}{\sqrt{2} } ∠ -60° ) * ( \frac{100}{\sqrt{2} } ∠ -30° )  

= 125 ∠ 30°

= ( 108.253 + 62.5 i ) VA   ( complex power )

active power = 108.25 watts

reactive power = 62.5 VAR'S

power factor = cos ∅ = cos 30° = 0.866 ( lagging )

current lags voltage by 30°

4 0
3 years ago
What information is needed to set up sales tax in QuickBooks Online for a client who only does business in their home state?
melisa1 [442]

Answer:

Their company address

When their last tax period started

How often they have to file a tax return

When they started collecting sales tax for the agency

Explanation:

For setup the sales tax information in Quickbooks online for a client who only does business in their home state, we need these information which are given below:

1. Their company addresses

2.  Last Tax Time period started

3.  How frequently they filed the tax return

3. when they begin to received sales tax

Therefore all the other options are not valid. Hence, ignored it

3 0
3 years ago
Blank is common during exercise.
dsp73

Answer:

Sweat

Explanation:

As you exercise you respire and warm up due to energy. In turn, two things happen, blood vessels vasodilate (irrelevant to you) and sweat glands sweat more. this sweat then evaporates and cools down the body.

7 0
2 years ago
1) Pareto charts are used to: A) identify inspection points in a process. B) outline production schedules. C) organize errors, p
zloy xaker [14]

Answer:

E) Please see below as the answer is self -explanatory.

Explanation:

The pareto chart, is used in quality control, and is a combined type of graph, that uses a line-type curve to denote the cumulative percentages of the different types of defects found in a sample (so the maximum value is 100%)

Also, it features a bar chart, which shows the relative occurrence of the different values (as in a histogram) which allows to find easily which defects are more relevant ones, alerting in this way about unacceptable deviations in the manufacturing process (if we are producing a good under given quality standards, for instance).

4 0
3 years ago
A 600 MW coal-fired power plant has an overall thermal efficiency of 38%. It is burning coal that has a heating value of 12,000
velikii [3]

Answer:

See step by step explanations for answer.

Explanation:

600 megawatts =

568 690.272 btu / second

thermal eficiency=work done/Heat supllied

0.38=568690.272/Heat supplied

Heat supplied=1496553.35btu /s

heat emmitted to the atmosphere=heat supplied -work done=(1496553.35-568690.272)=927863.1 btu/s

feed rate=(1496553.35)/12000=124.71 lb/s =10775184.1056 lb/day=5 387.472 ton / day

sulphur content released=(0.03*124.71)/(1.496553)=2.5 lb SO2/million Btu of heat input

so

the degree (%) of sulfur dioxide control needed to meet an emission standard=(2.5/0.15)*100=1666.67 %

the CO2 emission rate=220*(1.496553) =329.241 lb/s =12 903.0802 metric ton / day

5 0
3 years ago
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