Question

Water is flowing through a horizontal pipe of non-uniform cross-section. The speed of water is 30 cm/s
at a place where pressure is 10 cm (of water). Calculate the speed of water at the other place
where the pressure is half of that at the first place.​

Answer 1

Answer:

Explanation:

P = pressure

d = density of the liquid

v = velocity

h = height

g = accelerationdue to gravity

but question says that there is a horizontal pipe. so the  height will be zero.

so now  we can change above equation as below,

but we  pressure (p) = hdg

...(a)

substituing

h1g + (1/2)v1² = h2g + (1/2)v2²

10*10 + (1/2)*(30)² = 5*10 + (1/2)*(V2)²

100 + 450 = 50 + (v2²/2)

so now solving again

500 = v2²/2

1000 =v2²

31.62 cm/s =v2

speed of water = 31.62 cm/s plz mark brainliset if it deserve