contestada

10%) Problem 7: Water flows through a water hose at a rate of Q1 = 620 cm3/s, the diameter of the hose is d1 = 1.99 cm. A nozzle is attached to the water hose. The water leaves the nozzle at a velocity of v2 = 10.4 m/s. show answer No Attempt 17% Part (a) Enter an expression for the cross-sectional area of the hose, A1, in terms of its diameter, d1.

Respuesta :

MechEngineer

Answer:

a) [tex]A_1 = \frac{\pi d_1^2}{4}[/tex]

Explanation:

a) the cross-sectional area of the hose would be the square of radius times pi. And since the sectional radius is half of its diameter d. We can express the cross-sectional area A1 in term of diameter d1

[tex]A_1 = \pi r_1^2 = \pi (d_1/2)^2 = \frac{\pi d_1^2}{4}[/tex]

Otras preguntas