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3 years ago

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Given: Saturated air changes temperature by 0.5°C/100 m. The air is completely saturated at the dew point. The dew point has bee

n reached, and the temperature is 2°C. The air continues to rise another 1,400 m. What is the final temperature?

1 answer:

erma4kov [3.2K]3 years ago

3 0

Answer:

Explanation:

Given

saturated air temperature by $0.5^{\circ}C/100 m$

Dew point temperature is given by $t=2^{\circ}C$

Dew point is defined as the temperature after which air no longer to uphold the water vapor fuse with it and some water vapor may condense to a liquid.

air continues to rise for 1400 m

i.e. change in temperature would be $\Delta t =\frac{0.5}{100}\times 1400=7^{\circ}C$

Final temperature $t_f$

$t_f+\Delta t=t$

$t_f=2-7=-5^{\circ}C$

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Answer

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1. A perspex box has a 10 cm square base and contains water to a height of 10 cm. A piece of rock of mass 600g is lowered into t

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Answer:

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(c) The density of the rock is 30 g/cm³

Explanation:

The given parameters of the perspex box are;

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The initial level of water in the box, h₁ = 10 cm

The mass of the rock placed in the box, m = 600 g

The final level of water in the box, h₂ = 12 cm

(a) The volume of water in the box, 'V', is given as follows;

V = A × h₁

∴ The volume of water in the box, V = 10 cm² × 10 cm = 100 cm³

The volume of water in the box, V = 100 cm³

(b) When the rock is placed in the box the total volume, $V_T$ , is given by the sum of the rock, $V_r$ , and the water, V, is given as follows;

$V_T$ = $V_r$ + V

$V_T$ = A × h₂

∴ $V_T$ = 10 cm² × 12 cm = 120 cm³

The total volume, $V_T$ = 120 cm³

The volume of the rock, $V_r$ = $V_T$ - V

∴ $V_r$ = 120 cm³ - 100 cm³ = 20 cm³

The volume of the rock, $V_r$ = 20 cm³

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3 years ago

A 0.200-kg cube of ice (frozen water) is floating in glycerine.The gylcerine is in a tall cylinder that has inside radius 3.90 c

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Answer:

Part a)

h = 0.86 cm

Part b)

Level will increase

Explanation:

Part a)

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So here we will have

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Part b)

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