The height at which the atmosphere ceases to exist is about 80km. If the atmospheric pressure on the ground level is 760mmHg, the pressure at a height of 20km above the ground level is
(ρm = 13.6g/cm\(^3\) ρ = 0.00013g/cm\(^3\) )
380mmHg
570mmHg
190mmHg
480mmHg
Explanation
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If we assume that the atmospheric pressure decreases linearly with altitude, we can use the following formula:
\[ P = P_0 \cdot \left(1 - \frac{h}{H}\right) \]
Where:
- \( P \) is the pressure at the given altitude,
- \( P_0 \) is the pressure at sea level (ground level),
- \( h \) is the height above sea level (in this case, 20 km),
- \( H \) is the height at which the atmosphere ceases to exist (in this case, 80 km).
Given:
- \( P_0 = 760 \) mmHg,
- \( h = 20 \) km,
- \( H = 80 \) km.
Now, let's plug in the values into the formula:
\[ P = 760 \cdot \left(1 - \frac{20}{80}\right) \]
\[ P = 760 \cdot \left(1 - \frac{1}{4}\right) \]
\[ P = 760 \cdot \frac{3}{4} \]
\[ P = 570 \, \text{mmHg} \]
So, the pressure at a height of 20 km above the ground level is 570 mmHg.
Therefore, the correct answer is B. 570mmHg.
you can use the formula barometric formula.
p=p°(1-h/H)
where;
p°= atmospheric pressure at ground level
H= height at which the atmosphere ceases to exist
h= height where pressure is found
before I will finish solving this question, my system never off? eh my people...
May this question never come across us. absolute waste of time!!!. how am I supposed to know height of mercury = ATP-Pressure we're supposed to find
What is h in the calculation.
If height, y did in the substitution,: as in
d. h=d.h
h in his substitution is now pressure ( 760- p) mmHg,
How?
