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3 years ago

13

What happens to the boiling point of water as you climb higher up a mountain?

1 answer:

Maksim231197 [3]3 years ago

4 0

The higher the pressure, the higher boiling point of water. At lower the pressure, the boiling point of water comes down. So, the lower pressure inreases the boiling resulting more evaporation. As we go higher in altitude, the atmospheric pressure decreases. This results in decreasing the boiling point at higher altitude and increase in boiling of water. In fact, at the sea level ,the the sea water boils at 100 degree C where atmospheric pressre is normal. However , the boiling takes place at a lower temperature at the top of a mountain due to low pressure. In other words the boling is faster at the top of a mountain than that at its foot.

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A glass bottle of soda is sealed with a screw cap. The absolute pressure of the carbon dioxide inside the bottle is 1.50 x 105 P

MArishka [77]

Answer:

$F \approx 19.5 N$

Explanation:

From the question we are told that:

Pressure of $P_{CO_2}=1.50 * 105 Pa.$

Bottle cap area $A_b= 4.40 * 10-4 m^2$

Generally the equation for Resultant pressure $P_r$ is give as is mathematically given by

$P_r=P_{CO_2}-P_a$

Where

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2 years ago

The SI system uses three base units. Question 6 options: True False

GalinKa [24]

Answer:

The answer is false

Explanation:

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3 years ago

Consider a river flowing toward a lake at an average speed of 3 m/s at a rate of 550 m3/s at a location 58 m above the lake surf

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Answer:

1. 0.574 kJ/kg

2. 315.7 MW

Explanation:

1. The mechanical energy per unit mass of the river is given by:

$E_{m} = E_{k} + E_{p}$

$E_{m} = \frac{1}{2}v^{2} + gh$

Where:

Ek is the kinetic energy

Ep is the potential energy

v is the speed of the river = 3 m/s

g is the gravity = 9.81 m/s²

h is the height = 58 m

$E_{m} = \frac{1}{2}(3 m/s)^{2} + 9.81 m/s^{2}*58 m = 0.574 kJ/Kg$

Hence, the total mechanical energy of the river is 0.574 kJ/kg.

2. The power generation potential on the river is:

$P = m(t)E_{m} = \rho*V(t)*E_{m} = 1000 kg/m^{3}*550 m^{3}/s*0.574 kJ/kg = 315.7 MW$

Therefore, the power generation potential of the entire river is 315.7 MW.

I hope it helps you!

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