Question

An ideal fluid flows through a pipe made of two sections with diameters of 1.0 and 3.0 inches, respectively. The speed of the fluid flow through the 3.0-inch section will be what factor times that through the 1.0-inch section

Answer 1

`((r_1)/(r_2))^2=(1)/(9)`

Step-by-step explanation:

From the question we are told that:

Diameter 1
`d_1=1.0`

Diameter 2
`d_2=3.0`

Generally the equation for Radius is mathematically given by

At Diameter 1

`r_(1)=(1)/(2) inch`

At Diameter 2

`r_(2)=(3)/(2) inch`

Generally the equation for continuity is mathematically given by

`A_1V_1=A_2V_2`

Therefore

`((r_1)/(r_2))^2=((1/2)/(3/2))^2`

`((r_1)/(r_2))^2=(1)/(9)`