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3 years ago

Cual es la mayor diferencia entre un compuesto y una mezcla

Physics

2 answers:

Vanyuwa [196]3 years ago

6 0

Answer:

Un compuesto es cuando dos substancias se unen y deben ser quimicamente seperadas. Por ejemplo el elemento de hidrogeno y el elemento de oxigeno hacen agua. Una mezcla es cuando dos substancias se unen fisicamente for ejemplo una ensalada de lechuga y zanahorias.

ankoles [38]3 years ago

3 0

Answer:

Un elemento es un material compuesto de un simple tipo de átomo, un compuesto es una sustancia formada por dos o más elementos que se combinan químicamente y una mezcla es la combinación de sustancias, iguales o no, que pueden ser separadas por métodos físicos.

<em>espero que esto ayude :)</em>

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A ball is thrown vertically upward with a speed of 27.9 m/s from a height of 2.0 m. How long does it take to reach its highest p

Bas_tet [7]

Answer:

2.84403 seconds

2.91483 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-27.9}{-9.81}\\\Rightarrow t=2.84403\ s$

It takes 2.84403 seconds to reach the highest point

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-27.9^2}{2\times -9.81}\\\Rightarrow s=39.67431 m$

The ball will travel 39.67431+2 = 41.67431 m while going down to the ground

$s=ut+\frac{1}{2}at^2\\\Rightarrow 41.71479=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{41.67431\times 2}{9.81}}\\\Rightarrow t=2.91483\ s$

The ball takes 2.91483 seconds to hit the ground after it reaches its highest point.

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:

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

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3 years ago

Which lobes of the brain receive the input that enables you to feel someone scratching your back? (2 points)

andrew-mc [135]

Answer: Parietal

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3 years ago

How do you calculate acceleration

AfilCa [17]

A=f/m
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4 years ago

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Answer:

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4 years ago

Cual es la mayor diferencia entre un compuesto y una mezcla ​

Cual es la mayor diferencia entre un compuesto y una mezcla