A liquid (r 1.65 g/cm3) fl ows through two horizontal sections of tubing joined end to end. In the fi rst section, the cross-sectional area is 10.0 cm2, the fl ow speed is 275 cm/s, and the pressure is 1.20 105 Pa. In the second section, the cross-sectional area is 2.50 cm2. Calculate the smaller section’s

The question is incomplete: This is the complete Question

A liquid (ρ = 1.65 g/cm3) flows through two horizontal sections of

tubing joined end to end. In the first section the cross-sectional area is

10.0 cm2 , the flow speed is 275 cm/s, and the pressure is 1.20 x 105

Pa. In the second section the cross-sectional area is 2.50 cm2. Calculate

the smaller section’s (1) flow speed and (2) pressure.

(1) Smaller's section flow speed =

(2) Smaller's section pressure =

Step-by-step explanation:

Question 1

Using the continuity equation, we have that:

Question 2

Next, we apply Bernoulli's Equation

Let r = ρ(the density)

The pressure is given to us in its fundamental unit, Pa. So, we need to convert the speeds and density to their fundamental units.

≅

the answer for this question is : 360°

step-by-step explanation:

we have been given a triangle   we need to tell which one of these is a triangle below to map onto itself.

360° rotation is the complete rotation which eventually leads to the same figure which is given.

the correct will be, that is 360° that will cause the triangle below to map onto itself.