Water, which we can treat as ideal and incompressible, flows at 12 m/s in a horizontal pipe with a pressure of 3.0 x 10^4 pa. if the pipe widens to twice its original radius, what is the pressure in the wider section?

p2 = 9.8×10^4 Pa

Explanation:

Total pressure is constant and PT = P = 1/2×ρ×v^2  

So p1 + 1/2×ρ×(v1)^2 = p2 + 1/2×ρ×(v2)^2

from continuity we have ρ×A1×v1 = ρ×A2×v2  

v2 = v1×A1/A2  

and  

r2 = 2×r1

then:

A2 = 4×A1  

so,

v2 = (v1)/4  

then:

p2 = p1 + 1/2×ρ×(v1)^2 - 1/2×ρ×(v2)^2 = p1 + 1/2×ρ×(v1)^2 - 1/2×ρ×(v1/4)^2  

p2 = 3.0×10^4 Pa + 1/2×(1000 kg/m^3)×(12m/s)^2 - 1/2×(1000kg/m^3)×(12^2/16)  

     = 9.75×10^4 Pa

    = 9.8×10^4 Pa

Therefore, the pressure in the wider section is 9.8×10^4 Pa

the answer would be c. motion and direction

i hope this : )

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.