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?
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...