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2 years ago

47. Calculate the final steady temperature obtained when

Physics

1 answer:

Rashid [163]2 years ago

4 0

Answer:

the final steady temperature of the mixture is 46.67 ⁰C

Explanation:

Given;

mass of first water,m₁ = 6 g = 0.006 kg

initial temperature, t₁ = 50 ⁰C

mass of the second water, m₂ = 3.0 g = 0.003 kg

initial temperature of the second water, t₂ = 40 ⁰C

specific heat capacity of water, C = 4,200 J/kg.K

Let the final temperature of the mixture = T

Based on Law of conservation of energy, the temperature of the final mixture is calculated as follows;

m₁c(t₁ - T) = m₂c(T - t₂)

0.006 x 4200 x (50 - T) = 0.003 x 4200 x (T - 40)

1,260 - 25.2T = 12.6T - 504

1,260 + 504 = 12.6T + 25.2T

1764 = 37.8T

T = 1764/37.8

T = 46.67 ⁰C

Therefore, the final steady temperature of the mixture is 46.67 ⁰C

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Liula [17]

Answer:

7.5 cm

Explanation:

In the figure we can see a sketch of the problem. We know that at the bottom of the U-shaped tube the pressure is equal in both branches. Defining $\rho_A:$ Ethyl alcohol density and $\rho_G:$ Glycerin density , we can write:

$\rho_A\times g \times h_1 + \rho_G \times g \times h_2 = \rho_G \times g \times h_3$

Simplifying:

$\rho_A\times h_1 = \rho_G \times (h_3 - h_2) (1)$

On the other hand:

$h_1 + h_2 = \Delta h + h_3$

Rearranging:

$h_1 - \Delta h = h_3 - h_2 (2)$

Replacing (2) in (1):

$\rho_A\times h_1 = \rho_G \times (h_1 - \Delta h)$

Rearranging:

$\frac{h_1 \times (\rho_A - \rho_G)}{- \rho_G} = \Delta h$

Data:

$h_1 = 20 cm; \rho_A = 0.789 \frac{g}{cm^3}; \rho_G = 1.26 \frac{g}{cm^3}$

$\frac{20 cm \times (0.789 - 1.26) \frac{g}{cm^3}}{- 1.26\frac{g}{cm^3}} = \Delta h$

$7.5 cm = \Delta h$

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3 years ago

Which of the following states of matter is always a good conductor of electricity? solid, liquid, gas, or plasma **:/ Thank you!

pogonyaev

Plasma...I believe is always a good conductor of electricity. I was tempted to say a solid, but not all solids are the same in composition and that goes for liquid and gas as well.

Hopefully this helped and good luck.

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3 years ago

A 16000 kg railroad car travels along on a level frictionless track with a constant speed of 23.0 m/s. A 5400 kg additional load

Alisiya [41]

Answer:

The speed of the car when load is dropped in it is 17.19 m/s.

Explanation:

It is given that,

Mass of the railroad car, m₁ = 16000 kg

Speed of the railroad car, v₁ = 23 m/s

Mass of additional load, m₂ = 5400 kg

The additional load is dropped onto the car. Let v will be its speed. On applying the conservation of momentum as :

$m_1v_1=(m_1+m_2)v$

$v=\dfrac{m_1v_1}{m_1+m_2}$

$v=\dfrac{16000\ kg\times 23\ m/s}{(16000+5400)\ kg}$

v = 17.19 m/s

So, the speed of the car when load is dropped in it is 17.19 m/s. Hence, this is the required solution.

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Answer: Share

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