Water flows through a 2.5cm diameter pipe at a rate of
0.20m3/min, the absolute pressure in the pipe
being1.0x105 n/m. what must the diameter of
aconstriction in the pipe if the pressure at this point is to
be0.25x105
n/m?

From the Bernoulli energy principle,

ΔP + 1/2ρΔv² = 0 eqn 1

where

ΔP = pressure drop = P2 - P1 = (1 - 0.25)x10⁵ N/m =7.5 x 10⁴N/m

Δv²= velocity change = v₂² - v₁²

ρ = water density = 1kg/m3

Recall volumetric flow rate, Q=A v = constant

A = cross sectional area = πr²=πd²/4

d=pipe diameter at point 2 = 2.5cm = 0.025m and Q =0.20m³/min = 0.00333m³/s

So A= 0.000491m²

we can get v2 = Q/A = 6.79m/s

From eqn 1, v₁² = 2(P2 - P1)/ρ + v₂²

v₁² =  (2 x 7.5 x 10⁴)/1000 + 6.79²

v₁² = 196

v₁ = 14m/s

we can now get the area of the constriction point 2, A₁ = Q/v₁

A₁ = 0.000238m² and the diameter now will be d₁

d₁² = 4 x A₁ / π

     = 4 x 0.000238/3.14 = 0.000303m²

d₁ = √0.000303 = 0.0174m

Therefore, the diameter of  a constriction in the pipe  at the new pressure = 0.0174m = 1.74cm

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form

prndulum needs thread, weight clock to time swings