Recent content by bishoy

my apologies, I misunderstood. I thought you meant that J=kg*m^2/s^2. Your help is much appreciated :)

Well, When I plugged it into my calculator I acually used v = \sqrt { 2 \frac {\mbox{4.8x10^{-16}J}} {9.109x10^{-31}Kg } and as far as the units go, That was a good call... it Should have been m^2/s^2. Thank You for that. I really appreciate your...

alright so using this equation: v = \sqrt { 2 \frac {\mbox{KE}} m } and 9.10938188 × 10^(-31) kilograms as the mass of an electron i get v = \sqrt { 2 \frac {\mbox{4.8x10^{-16}J}} {9.10938188 × 10^{-31}Kg } 32463868.12 (assuming my math is...

Well, KE= 1/2mV^2, right? I have the KE, so all I need to do is put in the mass of the electron and solve for velocity? But I thought the mass of an electron is negligible in calculations?

Thanks Dick :) what about the second problem? any idea how to go about doing that one?

Homework Statement A bird stands on a DC electric transmission line carrying 2800A. The line has 2.5 x 10^-5 ohms resistance per meter, and the bird's feet are 4.0 cm apart. What is the potential difference between the bird's feet?Homework Equations I wish I knew The Attempt at a Solution...