Recent content by WINSTEW

yea I'm not getting that answer to work for the Efield at point (2,2.7)

I am lost again... Do i just put the u value ( x^2 +y^2) back in the equation and solve for x and then do the same thing but with ∂u/∂y for y? my x and y values were 2.0 and 2.7 respectively.

that symbol thing helps; so dV/du = -175*u^(-1.5) now understanding that its a ∂u/∂x with respect to x; =2x and with respect to y; =2y but how to i complete the equation to get an equation -175*x^(1.5)/2x and -175*y^(1.5)/2y for the x and y directions respectively?

ok so dV/dx = dV/du * du/dx dV = .5*350u^-.5 = 175u^-.5 du = 2x + 2y 175u^-.5/2x+2y * 2x+2y/2 =175u^-.5/2 for the x partial derivative and 175u^-.5/2 for the y partial doesn't look right

yea the equation 350/sqrt (x^2 +y^2) is given as what voltage equals and x and y are in meters. It then asks what the strength of the electric field is at x=2 and y = 2.7. I have to partially derive the equation. Once for x and once for y. Put the values in and solve for each component. I just...

so the problem is now I don't know how to partial derive the function. 350/ ( (x^2 + y^2)^-.5)

nevermind its -dv/dx which would give the -150 answer thanks... Finished that problem and now i have to do another one involving an x and a y component. Obviously you have to break the thing down into x and y but again I'm struggling on the partial derivative part. 350/(sqrt (x^2 + y^2))

so when i solve for x=1 i get 150. I put that answer in and it still says its wrong but to check my signs. I changed it to -150 and its correct... am i missing something?

Homework Statement The electric potential along the x-axis is V = 50x - (100/x)V, where x is in meters. What is E_x at x = 1.0m Homework Equations partial derivative of dV/dx, then plug in the value for x The Attempt at a Solution I understand the concept of trying to solve the...

Homework Statement Two objects are placed in a 1 dimensional box with springs of different constants at each end of the box. The box is made of wood and has a length of L. The first object has a mass of M and is initially compressed a distance of xinitial onto spring 1 of spring constant k...

yea i got down to that before.. Its definatly nasty but I still have 2 unknowns... 0= Lcos45 + Vinital*t + 4.905t^2. I know how to do the quadradic equation, but I still have V initial and T to solve. Is there anything else I can substitute in for V initial?

using trig the height of the ramp is Lcos45, so that would be my yinitial... but how do you find t in term of velocity?

ha ha definitely wrong...

wouldn't Vfy be 0? hitting the ground right?

ok I thought I knew how to do the first step but i have tried many times... To find the velocity at the top of the ramp you need to break it up into its components... First find the y velocity then the time then you can find the x velocity if you know the range (100m) and the time. I keep...