Recent content by lachy89

Speed is the magnitude of velocity and has no direction. EDIT: Assuming you are using the formula: v^2=u^2+2as you may need to check what you are using as v and u and be careful of your signs. 0 = u^2 +2as u^2=-2as Does that make sense?

You have not posted any relevant equations. I think a good starting point would be looking for an equation for the electric field due to a point charge. As you have explained the electric field is in the -x direction. Would this mean that there is no y component at the point in question?

http://en.m.wikipedia.org/wiki/Darcy_friction_fzactor_formulae Look at the equation section ( I realize it is wikipedia but it should stillbe correct). As the 'f' is on both sides of the equation you may need to do some iterations or look for a simplified solution.

The Equation for te Darcy friction factor is a 'rough formula' u believe to be more accurate and incorporate your e values look up the Colebrook white equation, this might change your f value. Ylt should be in m, I am a little unsure on the formula you used for this. Isn't there a v^2 meant...

Sorry you are correct I'm not sure why I wrote that bit on velocity. I wrote it from my phone so Iay not have proof read too well, sorry. Yes the viscosity only depends on the temperature. Just check the magnitude though it may be *10^-6 or something like that.

No the whole portion flows in both. So in the smaller pipe the area goes down, but to make the Q constant, the velocity goes up. Exactly ^^. Once again Q stays constant, so if A goes down V must go up. Your units are correct. m^3/s = m/s * m^2 (if that makes sense) Seems like an...

My quoting skills are lacking so I apologise. The volumes per second are the same in each pipe. If 0.0025m^3 per second is going in the left end then 0.0025m^3 must be coming out the right end. If this was not the case there would either be a build up of fluid in the pipe system somewhere...

There is no loss of water in the transfer of water from pipe 1 to pipe 2. So the volume flowing through one pipe must equal the volume flowing in the other pipe. The larger pipe has a larger supply for a constant velocity, but the velocity is not constant the volumetric flow is. If you...

Yes, you need the area of figure b. This I'd the cross sectional area. Your 1st reply included the surface area of a cylinder not the area of the circle in figure b. I would suggest converting to metres in all hydraulic problems (except maybe some special cases I cannot think of an example)...

The area in the equation Q=A*V is referring to the 'cross-sectional area' and so just the area of the circle. This is pi*r^2 or pi*d^2/4 which is used more often at the uni I attend for circular cross-sectional area. Also be careful to convert the diameters to metres. Lastly looking...

Hello, I am just an undergraduate studying Civil Engineering/Physics so take this as 'guidance' not complete help if you wish. From Understanding Hydraulics, 2nd ed., Hamill The equations I find of relevance are: Q=V*A Flow is equal to Velocity x Area Re = V*D/(v) Re =...

Ok I think I have seen where I have made the mistake, thank you for your 'guidance' :)

Eight 3.0 μC charges are located at the corners of a "unit cube centered about the origin with 1mm edges" r= ((1*10^-3)^2+(1*10^-3)^2+(1*10^-3)^2))^(1/2)m =1.73*10^-3m Wouldn't that be the distance?

Homework Statement Eight 3.0 μC charges are located at the corners of a unit cube centered about the origin with 1mm edges. How much work does it take to bring a 5.0 μC charge from infinity to the origin? Homework Equations U= k*q1*q2/r The Attempt at a Solution r=...

I'll show some of my working out in algebraic form so It will be easier to give some feedback. Torque = -p x E x sin(theta) Work = Torque x angle (in radians) Torque in this case is not constant as the torque changes with angle. Work = SUM OF ( Torque (theta) x d(theta)) from Pi/6 to...