Water flows through a water hose at a rate of q1 = 620 cm3/s, the diameter of the hose is d1 = 2.27 cm. a nozzle is attached to the water hose. the water leaves the nozzle at a velocity of v2 = 12.8 m/s.

a. enter an expression for the cross-sectional area of the hose, a1, in terms of its diameter, d1. sig. gif? tid=6c65-32-87-4c-b016-17686

b. calculate the numerical value of a1, in square centimeters.

c. sig. gif? tid=6c65-32-87-4c-b016-17686enter an expression for the speed of the water in the hose, v1, in terms of the volume flow rate q1 and cross-sectional area a1. sig. gif? tid=6c65-32-87-4c-b016-17686

d. calculate the speed of the water in the hose, v1 in meters per second

e. sig. gif? tid=6c65-32-87-4c-b016-17686enter an expression for the cross-sectional area of the nozzle, a2, in terms of v1, v2 and a1. sig. gif? tid=6c65-32-87-4c-b016-17686

f. calculate the cross-sectional area of the nozzle, a2 in square centimeters.

a) A₁ =  π d₁² / 4 , b) A₁ = 4.05 cm² , c) v₁ = Q / A₁ , d)  v₁ = 153 m / s , e)   A₂ = A₁ v₁ / v₂, f) A₂=  48.4 cm²

Explanation:

This is a fluid mechanics exercise, let's use the continuity equation

     Q = A₁ v₁ = A₂ v₂

Where Q is the flow, A are the areas and v the speeds

a) the area of the hose (A₁) that has a circular section is

     A₁ = π r₁²

Since the radius is half the diameter

    A₁ =  π (d₁ / 2)²

    A₁ =  π d₁² / 4

b) let's calculate

     A₁ =  π 2.27²/4

    A₁ = 4,047 cm²

    A₁ = 4.05 cm²

c) Let's use the left part of the initial equation

     Q = A₁ v₁

     v₁ = Q / A₁

d) let's calculate the value

    v₁ = 620 / 4.05

    v₁ = 153 m / s

e) We use the right part of the equation

    A₁ v₁ = A₂ v₂

    A₂ = A₁ v₁ / v₂

f) Calculate

A₂ = 4.05 153/12.8

A₂=  48.4 cm²

if you answer my question i will answer yours

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