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3 years ago

11

A 2,000-kg test car, traveling 60 m/s hits a brick wall. Using motion pictures, the time involved is determined to be 0.050 s. W

hat is the magnitude of the force that the car exerts on the wall?

1 answer:

IRISSAK [1]3 years ago

7 0

<h2>Answer</h2>

The force will be

$2.4 * 10^6 N$

<h2>Explanation</h2>

As given in question

Mass= 2000 kg

Velocity= 60 m/s

Time = 0.05 sec

As we know

$Acceleration = Velocity/Time$

$Acceleration = 60 / 0.05$

$Acceleration = 1200 ms^{-2}$

For force we know that

$Force= Mass * Acceleration$

$Force= 2000 x 1200$

$Force = 2400000 N$

$Force = 2.4 * 10^6 N$

So, The answer is

$Force = 2.4 * 10^6 N$

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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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3 years ago

WILL GIVE BRAINLIEST TO THE CORRECT ANSWER

LenKa [72]

Answer:

See the answers below.

Explanation:

We can solve both problems using Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

∑F =m*a

where:

F = force [N] (units of newtons)

m = mass = 1000 [kg]

a = acceleration = 3 [m/s²]

$F = 1000*3\\F=3000[N]$

And the weight of any body can be calculated by means of the mass product by gravitational acceleration.

$W=m*g\\W=1000*9.81\\W=9810 [N]$

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