Recent content by ostrik23

I do agree it seems like an awfully small force, but the problem was given as such. As for the result, it heavily makes me doubt my answer in a), given that 0.08 Newtons is able to empty the syringe without a needle in 10 seconds, whereas .1 Newton won't be able to empty the same amount of...

Okay, so volumetric flowrate Q=\frac{\pi r^4 \Delta p}{8\mu L}=\frac{\pi \cdot (0.105\cdot 10^{-3})^4\cdot 503}{8\cdot 10^{-3}\cdot 25\cdot 10^{-3}}=9.603\cdot 10^{-10} Then t=\frac{V}{9.603\cdot 10^{-10}}=\frac{10^{-5}}{9.603\cdot 10^{-10}}=10413 Which still seems like an unreasonably long...

I am guessing the correct units are Pascal Seconds? (Pas) Even so, if I calculate my value for t with the expression I found t=\frac{8\cdot 10^{-3}\cdot 25\cdot 10^{-3}\cdot 10^{-5}}{\pi \left(\frac{15.9}{2}\cdot 10^{-3}\right)^4\cdot 503}=3.17\cdot 10^{-4} Which is way too fast. I just don't...

Okay so for parameter values I used \mu = 10^{-3} \textrm{pascal} and for radii i used r=\frac{0.210}{2} \cdot 10^{-3} \textrm{m} For L i used the length of the needle L=25\cdot 10^{-3} \textrm{m} Where it gets a bit sketchy is in my calculation of \Delta...

Okay, so what I attempted for a) is to use Bernoulli's. The velocity of the fluid in the chamber should be equal to the velocity of the piston, which in comparison to the barrel should be negligible. Hence I get \frac{p_1}{\rho g} = \frac{1}{2} \frac{v_2^2}{2} + \frac{p_2}{\rho g} so \Delta...