Water, with a density of rho = 1145 kg/m 3 , flows in a horizontal pipe. In one segment of the pipe, the flow speed is v 1 = 7.33 m/s . In a second segment, the flow speed is v 2 = 1.57 m/s . What is the difference between the pressure in the second segment ( P 2 ) and the pressure in the first segment ( P 1 )?
Answers
Answer:
the pressure difference will be ΔP= P₂ - P₁ =29348.64 Pa
Explanation:
from Bernoulli's equation
P₁ + ρgh₁+ 1/2ρv₁²= P₂ + ρgh₂+1/2ρv₂²
where P = pressure , ρ= density , g= gravity , h= height , v=flow speed and 1 and 2 denote first and second segment respectively
then since the pipe is horizontal there is no difference in height (h=h₁=h₂) , thus
P₁ + 1/2ρv₁²= P₂ + 1/2ρv₂²
the pressure difference ΔP= P₂ - P₁ will be
ΔP= P₂ - P₁ = 1/2ρv₁² - 1/2ρv₂² = 1/2*ρ* (v₁² - v₂²)
replacing values
ΔP= 1/2*ρ* (v₁² - v₂²) = 1/2 *1145 kg/m³ * [(7.33 m/s)² - (1.57 m/s)²] = 29348.64 Pa
ΔP= 29348.64 Pa
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Answers
Answers
Answer:
The answer is 2.71 g/mL
Explanation:
The density of a substance can be found by using the formula
From the question
mass = 244 g
volume = 90 mL
So we have
We have the final answer as
2.71 g/mL
Hope this helps you