Au-tube open at both ends is partially filled with water (fig. p14.81a). oil having a density 750 kg/m3 is then poured into the right arm and forms a column l 5 5.00 cm high (fig. p14.81b). (a) determine the difference h in the heights of the two liquid surfaces. (b) the right arm is then shielded from any air motion while air is blown across the top of the left arm until the surfaces of the two liquids are at the same height (fig. p14.81c). determine the speed of the air being

1. 4042.5

2. 5.687 m>2

Explanation:

Question 1: pressure difference = height × density × g ,   p = h × p × g

Question 2: Δ(K.E.)=12(ρA 1v1Δt)(v22−v21).  

The only possible source for this increase in energy is the work done by pressure in pushing the fluid into the upward

Taking the pressure on area  A1  to be  P1,  the total force on  A1  is  P1A 1.   In the time  Δt,  this force acts through a distance  v1Δt,  and hence does work =  force × distance =P1A1v1Δt. .  

So this is work done on our chunk of fluid by the fluid pushing it from behind — but that’s not the end of the story, because our chunk of fluid itself does work pushing the fluid in front of it, so to find the total increase in our chunk’s energy, we must subtract off the external work it does.  That is, the total work done by pressure on our fluid is

P1A 1v1Δt−P2A2v2Δt=(P1−P2)A 1v1Δt  

remembering that  A 1v1=A2v2.  

This work done must equal the change in kinetic energy, so

(P1−P2)A 1v1Δt=12(ρA 1v1Δt)(v∧2−v2∧1)  

from which  

P1+12ρv21=P2+12ρv22.

p1-p2=1/2p2-p1(v2-v1)2

4042.5=1/2*100-750(v2-v1)

8085/250=(v2-v1)

32.34=(v2-v1)2

v2-v1=32.34

the electric field is a “force field” around a charged object that illustrates the direction the electric force would push an imaginary 

did yall get answers for this yet