Recent content by CollegeStudent

Oh man...That was just a blonde moment...it took me actually calculating it out to see that it is just atmospheric pressure. That was fun haha!

Perfect! Also, 1 LAST question that comes from the Boyle's Law relation we came up with earlier P_{f} = P_{i}\frac{V_i}{V_f} Finding the initial pressure inside of the pressure chamber, using the ideal gas law, leads to P_{i} = \frac{mRT}{V} Which, when replacing m = \rho V leads to P_{i} =...

Hey billy_joule, Sorry to bother you again, but any more insight on this? While before, we could have calculated the velocity leaving the nozzle *ignoring losses* Now, taking into account the losses, we have 2 more terms both requiring a velocity h_{l} = f \frac{L}{D} \frac{V^2 }{2} and h_{ml}...

Hmm, although, the Reynolds number... Re = \frac{VD}{v} since I need the velocity FOR the Reynolds number...maybe it isn't quite as easy as I am thinking?

Absolutely! And I do plan on having this losses accounted for! Major head losses = h_{l} = f \frac{L}{D} \frac{V^2}{2} Where 'f' is the friction factor found via the Reynolds number and the roughness Minor head losses = h_{ml} = k \frac{V^2}{2} where k will depend on the bends in the...

Perfect, and that would be the pressure built up in the chamber after corresponding pumps Now, if I wanted to find the velocity of water leaving the nozzle when the trigger is pushed...since I know the pressure in the chamber, I could set up a Bernoulli relation correct...

Got it, so to work with a general example... let's say the volume of the chamber is 5000in^3 originally when filled with just air. The *cylindrical* piston has an 11 [in] stroke and we'll say a 2 [in] diameter...so the volume would be that same 2265 in^3 This volume would effectively be moved...

Ha, obviously I would need to change the volume of the chamber...that's what I get for choosing random numbers! Okay let's instead say the Volume of the Pressure Chamber is 5000 in^3 *again just choosing random numbers, as I know this is unrealistic for this application. This is where I get a...

Alright, so let's see...lets take this animation for example: http://static.howstuffworks.com/flash/water-blaster-soaker.swf ****Only difference will be that originally the only water will be in the reservoir, nothing in the system**** Lets say the stroke length is 11 [in.] and the cylinder the...

Hello all, I have a question about a water gun problem I'm working on. The general question is...how can you associate the hand pump stroke length, and the diameter of the pipe the pump is inside, with how much pressure gets built up in the guns pressure chamber?

1. Homework Statement Determine whether the block shown is in equilibrium and find the magnitude and direction of the friction force when θ = 25° and P = 750N μs[\SUB] = .35 μk[\SUB] = .25 2. Homework Equations Fs = μs * N Fk = μk * N 3. The Attempt at a Solution Attachment 1 up there...

Oh wow was that a mistake...what got me confused on that was I was looking at it as (1/5) (.2Q / (pi * r)) which then lead me to the (5 pi * r) alright so then it SHOULD be ,if wanting to use decimals here, (.2pi * r) Well that's interesting...now the (.2)'s cancel and we only have the...

Alright so I've been working on this...and understand that this question want the field at the center of the circle So here's what I've been doing... Total Charge = Q Charge per unit length = Q/L...

Doesn't specify...That is the entire question...hmm based on the level of what the class would be at, I would say at some general point

Homework Statement Find the electric field due to a half circle of radius R and total charge Q. (a) First estimate by dividing the charge into five equal parts. (b) Repeat for 9 equal parts, (c) Using integral calculus solve exactly.Homework Equations E = kQ/r^2 The Attempt at a Solution...