Recent content by jimithing

I got F = \frac{q^2}{4\pi \epsilon_{0}(3d)^2} , F = -k(3d) So k = -\frac{q^2}{4\pi \epsilon_{0}27d^3} Do I now just sub into U = \frac{1}{2}k(3d)^2 ?

A light unstressed spring has length d. Two identical particles, each with charge q, are connected to the opposite ends of the spring. The particles are held stationary a distance d apart and then released at the same time. The system then oscillates on a horizontal frictionless table. The...

i assumed that the 25% and 12% mixtures were sold at cost, so i used .25x + .75y = 18 and .12x + .88y = 10 solved for x and y, then substituted into .17x + .83y = z found z, multiplied by 1.1 hopefully that's what they're looking for

A paint mixture containing 25.0% pigment and the balance water sells for $18.00/kg, and a mixture containing 12.0% pigment sells for $10.00/kg. If a paint retailer produces a blend containing 17.0% pigment, for how much ($/kg) should it be sold to yield a 10% profit? Am I wrong, or are they...

A long coaxial cable consists of an inner cylindrical conductor with radius a and an outer cylindrical shell of inner radius b and outer radius c. The cylindrical shell is mounted on insulating supports and has no net charge. The inner cylinder has a uniform positive charge per unit length...

i figured it out earlier today F = ma qE = ma \frac{eE}{m}= a find a, use it to find the displacement (x) in the simple mechanics equation \frac{-v_{0}^2}{2a}= x then use x to find t : \sqrt{\frac{2x}{a}}= t now simply use: \Delta KE = \frac{1}{2}mv_{0}^2 -...

another hint - the electrical field outside a spherical shell of charge with radius r and charge q is directed radially and has magnitude E = \frac{q}{4\pi \epsilon_{0}r^2}

use a gaussian surface and take advantage of its symmetry, but make sure your point P is on the GS or else there will be error. \Phi = \oint E \cdot dA E will be constant, so it can be taken out of the integration, etc.

The viewing window in a diving suit has an area of 65 cm^2. If an attempt were made to maintain the pressure on the inside of the suit of 1 atm, what force would the window have to withstand if the diver descended to a depth of 150 m. Take the specific gravity of the water to be 1.05. So...

just use vector addition for the direction (head to tail addition)

the thing is, is that we haven't studied potential...i was trying to work this problem using strictly kinematics...

figured it out in class today, thanks

try 120 degrees for the second batch of steps (60 degrees north of west)

could you define u, s, V, W, and Q

here can \vec {a} be represented by \frac{v^2 - v_o^2}{2(x - x_o)} , with v being the velocity given?