Recent content by Adam Quinn

okay so, {A_1}{V_1} = {A_2}{V_2} {A_1} = {2.14*10^{-5}} \dfrac{A_1}{A_2} = \dfrac{V_2}{V_1} \dfrac{2.14*10^{-5}}{A_2} = \dfrac{V_2}{V_1} what do I do from here, please no more riddles, struggling to understand this as it is.

hence {2.14*10^{-5}}*{v_1}={A_2}*{12.78}

this would essentially mean that {v_2} = 12.779 m.s^{-1}

By solving this I get a negative velocity value, something tells me this means that it's incorrect.

This could be expressed as {v_2} = {2.18*10^{-5}}*{v_1} If this is true then {182976}-{101325} = {500}*(({2.18*10^{-5}*v_1})-{v_1})

hence, \dfrac{A_1}{2.18*10^{-5}} = \dfrac{v_1}{v_2} where A1 would be negligible? hence \dfrac{v_1}{v_2} = 2.18*10^{-5}

I'm failing to see how this helps me, I can't just say that the cross sectional area of the needle is 0, nor can I say that the velocity is extremely large. (as per a1v1=a2v2, I don't even have a velocity for the tube.)

if not then i have (182976 - 101325)/.5*1000 = v2^2-v1^2

p2 doesn't increase from 1atm when the plunger is pushed?

Yes, I have P1, cannot find P2, I can eliminate pgh from both sides of the equation. I am now left with P1 = 182976.376, A1 = 2.18*10^-5, and the rest of the equation p1-p2 = 1/2rho (v^2 - v^2)

uh, how? I really don't know...

for study

If it says that the pressure in the needle remains at 1atm, even though liquid is being pushed through, would the force through the first part be equal to the force equal through the needle, and hence p1a1=p2a1? Would this give me a correct area value?

I wasnt given any other measurements, how can I solve for the area of the needle?

A hypodermic syringe contains a medicine with the density of water (figure below). The barrel of the syringe has a cross-sectional area of 2.18 10-5 m2. In the absence of a force on the plunger, the pressure everywhere is 1.00 atm. A force of magnitude 1.78 N is exerted on the plunger, making...