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kakasveta [241]

4 years ago

13

If a freely falling object were equipped with a speedometer on a planet where the acceleration due to gravity is 20 m/s², then i

ts speed reading would increase each second by If a freely falling object were equipped with a speedometer on a planet where the acceleration due to gravity is 20 m/s², then its speed reading would increase each second by __________.a. 10 m/s. b. 30 m/s. c. 20 m/s. d. 40 m/s. e. depends on its initial speed

2 answers:

babymother [125]4 years ago

7 0

Answer:

20 m/s

Explanation:

The acceleration of an object determines the increase in velocity per second.

In this case the acceleration is 20m/s^2.

$a=\frac{v}{t}=\frac{\frac{20m}{s}}{s}$

That is, the increase is 20m/s each second.

I hope this is useful for you

Regards

DanielleElmas [232]4 years ago

7 0

Answer:

velocity will increase 20 m/sec each second

Explanation:

Given that the acceleration due gravity is $g=20m/sec^2$

We have to find the increase in the speed each second

We know that acceleration is nothing but rate of change of velocity as it changes $20m/sec^2$ so the speed will also changes with 20 m/sec

the object is falling freely so initial velocity u =0 m/sec

Acceleration due to gravity g= 20 m/s²

Time = 1 sec

From first law of motion

$v=u+gt\\=0+20\times 1\\=20m/sec$

the velocity will increase 20 m/sec each second

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3 years ago

A bin is given a push across a horizontal surface. The bin has a mass m, the push gives it an initial speed of 1.60 m/s, and the

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Answer:

The bin moves 0.87 m before it stops.

Explanation:

If we analyze the situation and apply the law of conservation of energy to this case, we get:

Energy Dissipated through Friction = Change in Kinetic Energy of Bin (Loss)

F d = (0.5)(m)(Vi² - Vf²)

where,

F = Frictional Force = μR

but, R = Normal Reaction = Weight of Bin = mg

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where,

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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;

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

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

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

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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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