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nataly862011 [7]
3 years ago
11

The barometric pressure at sea level is 30 in of mercury when that on a mountain top is 29 in. If the specific weight of air is

assumed constant at 0.0075 lb/ft3 , calculate the elevation of the mountain top.
Physics
1 answer:
stealth61 [152]3 years ago
7 0

To solve this problem we will apply the concepts related to pressure, depending on the product between the density of the fluid, the gravity and the depth / height at which it is located.

For mercury, density, gravity and height are defined as

\rho_m = 846lb/ft^3

g = 32.17405ft/s^2

h_1 = 1in = \frac{1}{12} ft

For the air the defined properties would be

\rho_a = 0.0075lb/ft^3

g = 32.17405ft/s^2

h_2 = ?

We have for equilibrium that

\text{Pressure change in Air}=\text{Pressure change in Mercury}

\rho_m g h_1 = \rho_a g h_2

Replacing,

(846)(32.17405)(\frac{1}{12}) = (0.0075)(32.17405)(h_2)

Rearranging to find h_2

h_2 = \frac{(846)(32.17405)(\frac{1}{12}) }{(0.0075)(32.17405)}

h = 9400ft

Therefore the elevation of the mountain top is 9400ft

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