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 )?

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

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2