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3 years ago

Calculate the kinetic energy in joules of a 1,500 kg car that is moving at a speed of 42 km/h

1 answer:

4 0

Data:
KE (Kinetic Energy) = ? (Joule)
m (mass) = 1500 Kg
v (speed) = 42 Km/h
converting to m/s (42 / 3.6), we have: v (speed) = 11.6 m/s

Formula:
$K_{E} = \frac{1}{2} m*v^2$

Solving:
$K_{E} = \frac{1}{2} m*v^2$
$K_{E} = \frac{1}{2} *1500*(11.6)^2$
$K_{E} = \frac{1}{2} *1500*134.56$
$K_{E} = \frac{201840}{2}$
$\boxed{\boxed{K_{E} = 100920\:Joule}}\end{array}}\qquad\quad\checkmark$

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If the solar system shrank so that the sun were located just one centimeter from Earth, about how far away could you find the ne

DedPeter [7]

E S *

The "E" represents Earth, "S" represent Sun, and the "*" represents the nearest star(which is Proxima Centauri).

The main thing to worry about here is units, so ill label everything out.
D'e,s'(Distance between earth and sun) = .00001581 light years
D'e,*'(Distance between earth and Proxima) = 4.243 light years

Now this is where it gets fun, we need to put all the light years into centimeters.(theres alot)
In one light year, there are 9.461 * 10^17 centimeters.(the * in this case means multiplication) or 946,100,000,000,000,000 centimeters.

To convert we multiply the light years we found by the big number.
D'e,s'(Distance between earth and sun) = 1.496 * 10^13 centimeters
D'e,*'(Distance between earth and Proxima) = 4.014 * 10^18 centimeters

Now we scale things down, we treat 1.496 * 10^13 centimeters as a SINGLE centimeter, because that's the distance between the earth and the sun. So all we have to do is divide (4.014 * 10^18 ) by (1.496 * 10^13 ).
Why? because that how proportions work.

As a result, you get a mere 268335.7 centimeters.

To put that into perspective, that's only about 1.7 miles

A lot of my numbers came from google, so they are estimations and are not perfect, but its hard to be on really large scales.

8 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:

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

3 0

3 years ago

What happens if someone walks between you and the heater? Can you still feel its warmth?

Pie

No, for that moment

6 0

4 years ago

Read 2 more answers

10 kg cart and a 5 kg cart are placed on identical surfaces. The 10 kg cart experiences a net force of 12 N to the left, while t

SashulF [63]

F=ma

For the first (10kg) cart,
12=10a
a=6/5 m/s^2 to the left

For the second (5kg) cart,
8=5a
a=8/5 m/s^2 to the left

Therefore, the lighter (5kg) cart experiences a greater acceleration.

7 0

3 years ago

147.5 ml of ethyl alcohol (coefficient of volume expansion = 1120 x 10-6K-1) at a temperature of 273.1 K is measured into a 150.

marishachu [46]

Answer:

288.2 K

Explanation:

$\beta$ = Volumetric expansion coefficient = $1120\times 10^{-6}\ /K$

$T_i$ = Initial temperature = 273.1 K

$T_f$ = Final temperature

$v_0$ = Original volume = 150 mL

Change in volume is given by

$\Delta v=\beta v_0(T_f-T_i)\\\Rightarrow T_f=\frac{\Delta v}{\beta v_0}+T_i\\\Rightarrow T_f=\frac{150-147.5}{1120\times 10^{-6}\times 147.5}+273.1\\\Rightarrow T_f=288.2\ K$

The temperature of the ethyl alcohol should be 288.2 K to reach 150 mL

8 0

4 years ago