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2 years ago

How many joules are there in 200kj

Physics

2 answers:

timurjin [86]2 years ago

5 0

Answer:

$2.00 * 10^5J$

Explanation:

Make use of your conversion factors:

$200KJ * \frac{1000J}{1kJ} = 200,000J = 2.00* 10 ^5J$

Or alternatively make use of your scientific notation. $kilo = 10^3$

Therefore $200kJ = 200*10^3J=2.0*10^2*10^3J = 2.0*10^5J$

Lesechka [4]2 years ago

4 0

Answer:

200,000 joules <3

Explanation:
just accept it

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The starships of the Solar Federation are marked with the symbol of the Federation, a circle, whereas starships of the Denebian

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Complete question

The complete question is shown on the first uploaded image

Answer:

The velocity is $v = c* \sqrt{1 - \frac{1}{n^2} }$

Explanation:

From the question we are told that

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Now this length seen by the observer can be mathematically represented as

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i.e h = b

$b = nb [\sqrt{1 - \frac{v^2}{c^2} } ]$

$[\frac{1}{n} ]^2 = 1 - \frac{v^2}{c^2}$

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6 0

3 years ago

What is your operational definition for a fast reaction time?

LuckyWell [14K]

Answer:

The ability to react to a certain stimulus with a speedy and effective manner

Explanation:

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3 years ago

An ocean thermal energy conversion system is being proposed for electric power generation. Such a system is based on the standar

defon

Answer:

Explanation:

Dear Student, this question is incomplete, and to attempt this question, we have attached the complete copy of the question in the image below. Please, Kindly refer to it when going through the solution to the question.

To objective is to find the:

(i) required heat exchanger area.

(ii) flow rate to be maintained in the evaporator.

Given that:

water temperature = 300 K

At a reasonable depth, the water is cold and its temperature = 280 K

The power output W = 2 MW

Efficiency $\zeta$ = 3%

where;

$\zeta = \dfrac{W_{out}}{Q_{supplied }}$

$Q_{supplied } = \dfrac{2}{0.03} \ MW$

$Q_{supplied } = 66.66 \ MW$

However, from the evaporator, the heat transfer Q can be determined by using the formula:

Q = UA(L MTD)

where;

$LMTD = \dfrac{\Delta T_1 - \Delta T_2}{In (\dfrac{\Delta T_1}{\Delta T_2} )}$

Also;

$\Delta T_1 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_1 = 300 -290 \\ \\ \Delta T_1 = 10 \ K$

$\Delta T_2 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_2 = 292 -290 \\ \\ \Delta T_2 = 2\ K$

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$LMTD = \dfrac{8}{In (5)}$

LMTD = 4.97

Thus, the required heat exchanger area A is calculated by using the formula:

$Q_H = UA (LMTD)$

where;

U = overall heat coefficient given as 1200 W/m².K

$66.667 \times 10^6 = 1200 \times A \times 4.97 \\ \\ A= \dfrac{66.667 \times 10^6}{1200 \times 4.97} \\ \\ \mathbf{A = 11178.236 \ m^2}$

The mass flow rate:

$Q_{H} = mC_p(T_{in} -T_{out} ) \\ \\ 66.667 \times 10^6= m \times 4.18 (300 -292) \\ \\ m = \dfrac{ 66.667 \times 10^6}{4.18 \times 8} \\ \\ \mathbf{m = 1993630.383 \ kg/s}$