Recent content by MD2000

I figured on top V2=.5V1 (because it has twice the area) Then... P1 + DGH + .5DV^2 = P2 + DGH + .5DV^2 I figured P would need to be equal so that cancels and then .5V1 could be plugged in for V2 and solve for H? Honestly I'm not sure

Ahh..i see..thanks.. can anyone help me with the other q?

Ok yea..they are moving towards each other both at v...would it really make a difference if both or one were moving? Is that 2 x .5mv^2 really necessary?

Anyone?

I have two questions that I want to make sure I'm doing correctly and understanding. The first question is how close will two protons get to each other if they are traveling at a velocity v towards each other. I figured you can set up 1/2mv^2 = kq1q2/r and solve for r..is that correct...

In a Young's double-slit experiment, the seventh dark fringe is located 0.023 m to the side of the central bright finge on a flat screen, which is 1.1 m away from the slits. The separation between the slits is 1.5 10-4 m. What is the wavelength of the light being used? m I'm having some...

How exactly can I get the currents for each individual peice? I know that the total current is going to be V = IR..which equals 2.46.. But how can I solve for individual currents? Since you don't have V..what can you use to get the I?

Determine the power dissipated in the 6.0 resistor in the circuit shown in the drawing. (R1 = 4.0 , R2 = 6.0 and V1 = 15 V.) Am I supposed to be using P = V^2/R..is the power flowing through each resistor going to be diff? Anyone want to help me out where to start?

Ahh..I completely forgot the negative..thanks astronuc

A metal sphere has a charge of +5.0 µC. What is the net charge after 7.0 x 10^13 electrons have been placed on it? What I tried was to convert the number of electrons to a charge by multiplying it by (1.6 x 10^-19)..and then adding that to the 5..from which I got 16.2..am I takin the right...

Hmm..so I've tried a couple different methods and am getting a huuge answer..not sure what I'm doing wrong..any help with this or the third one anyone?

hmm..when I solve for T I'm getting a huge number.. MCT (water) + MCT(Calo.) + MLV (steam) = MCT(Copper) (.203)(4186)(38-16) + (.150)(387)(38-16) + (.2018)(22.4x10^5) = (.093)(387)(38-Tf) any idea what I am doing wrong?

Ahh..gotcha..thanks.. what about the third one? Did I simply ratio it wrong? Is it 4 instead of .25?

Hmm.. MCT (water) + MCT(Calo.) + MCT(steam) MLV (steam) = MCT(Copper) (.203)(4186)(38-16) + (.150)(387)(38-16) + (.2018)(1970)(38-16) + (.2018)(33.5x10^4) = (.093)(387)(38-Tf) And simply solving for Tf?

Ahh..I always make stupid mistakes..so since its vaporizing..I need to use mLv on the right side of the equation to account for the vaporization of the fluid?