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3 years ago

Rihanna can go from 0-60 miles per hour in 3.5 seconds. Calculate the acceleration.

Physics

1 answer:

Anton [14]3 years ago

8 0

Answer:

7.66 m/s²

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 0 mph

Final velocity (v) = 60 mph

Time (t) = 3.5 s

Acceleration (a) =.?

Next, we shall convert mile per hour (mph) to metre per second (m/s). This can be obtained as follow:

Recall:

1 mph = 0.447 m/s

Initial velocity (u) = 0 mph

Initial velocity (u) = 0 × 0.447 = 0 m/s

Final velocity (v) = 60 mph

Final velocity (v) = 60 × 0.447 = 26.82 m/s.

Finally, we shall determine the acceleration of Rihanna as follow:

Initial velocity (u) = 0 m/s

Final velocity (v) = 26.82 m/s.

Time (t) = 3.5 s

Acceleration (a) =.?

a = (v – u) /t

a = (26.82 – 0) / 3.5

a = 26.82 / 3.5

a = 7.66 m/s²

Thus, the acceleration of Rihanna is 7.66 m/s²

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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;

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

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

$LMTD = \dfrac{10 -2}{In (\dfrac{10}{2} )}$

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

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:

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

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