Bernouli's Equation/Pressure Problem

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SUMMARY

The discussion centers on applying Bernoulli's Equation to determine the pressure difference between the inside and outside of a truck's tarpaulin while traveling at 27 m/s. The air density is given as 1.29 kg/m³. By substituting the known values into the equation, the pressure difference is calculated as 470 Pa, confirming that the pressure inside the cargo area exceeds the outside pressure by this amount. The solution effectively demonstrates the application of Bernoulli's principle in a real-world scenario involving fluid dynamics.

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Homework Statement


Consider a truck with a tarpaulin on top of it. When the truck is stationary the tarpaulin lies flat, but it bulges outward when the truck is speeding down the highway. The truck is traveling at 27 m/s. The density of air is 1.29 kg/m^3. by how much does the pressure inside the cargo area beneath the tarpaulin exceed the outside pressure

Homework Equations


bernoulis:
P(1) + 1/2pv^2 = P(2) + 1/2pv^2


The Attempt at a Solution


i plugged in the density and but with two unknown pressures and one unknown velocity i don't know how you would find the difference
 
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and i think the pgy's on either side of the equation cancel out because this is horizontal, so y=0.
 
I would consider the air inside the truck (under the tarpaulin) as stationary.
You can imagine the truck at rest and someone blowing some air on the outside.
 
When the truck is traveling at 27 m/s, the air above the tarpaulin is moving with the same velocity but in the opposite direction. The pressure inside this blowing air decreases. In side the tarpaulin air is at rest.
 
so if the air inside the tarpaulin can be considered stationary, you can use P(2) = P(1) + pgh. And the pgh would = 0, so the pressure inside would equal 1 x 10^5 Pa?
 
AH oh wait is this how you solve it:

P(1) + 1/2pv^2 = P(2) + 1/2pv^2
P(1) - P(2) = 1/2p(v2^2 - v1^2)
P(1) - P(2) = 1/2(1.29)(27^2 - 0^2)
P(1) - P(2) = 470 Pacorrect??
 
Correct.
 
thanks profusely
 

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