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3 years ago

What do our ears do

2 answers:

8 0

Answer;

Hear

Explanation:

Your ears do the remarkable job of allowing you to hear a huge range of sounds, from a whisper to a loud bang. To do this, the ear transforms sound energy into electrical signals which the brain can interpret. Your ears also help to maintain your balance.

Can I get brainliest

andre [41]3 years ago

7 0

Answer:

hear

Explanation:

stuff stuff moaning and stuff

You might be interested in

A Jamaican Bobsled team hits the brakes on their sled so that decelerates at a uniform rate of 0.43 m/s^2. How long does it take

ankoles [38]

Answer:

20.05 seconds

Explanation:

Given that:

v² = u² + 2as

v = final velocity

u = initial velocity

a = acceleration

s = distance

a = negative acceleration = - 0.43

s = distance = 86.5

v = 0

0 - 2(0.43)(86.5) = u²

74.39 = v²

U = sqrt(74.39)

U = 8.6249

From ;

t = Time in seconds

t = (v - u) / a

t = (0 - 8.6249) / - 0.43

t = 20.057

6 0

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;

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

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

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

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

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:

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