Okay, I will try to figure it out. In the meantime, can you help me with another question?
The problem is as follows:
In nuclear fission, a nucleus splits roughly in half.
(a) What is the potential 3.50 10-14 m from a fragment that has 50 protons in it?
I calculated this to be...
1.)1/2m(Vf^2-Vi^2) + q(Vf-Vi)=0, cancel out initial velocity and final voltage (I think) since they are zero.
2.)1/2mVf^2-qV=0
3.) 1/2mVf^2=qV
4.)1/2mVf^2/q=V
That's what I tried, and got the wrong answer so I guess that I am setting up the equation wrong.
Ha ha okay, I actually tried doing that earlier but I got the wrong answer. Which of the Velocities and voltages are zero? And for the mass of the electron do I use 9.1X10^-31?
If the electron travels 0.35 m in 46 ns, what velocity does it travel at? Hence what KE does it posses at this velocity? Also how does voltage and charge relate to energy?
Well the velocity is going to be at about 7.6x10^6 m/s. I know that K.E.= 1/2mv^2, so can I calculate the kinetic energy...
Homework Statement
If an electron travels 0.350 m from an electron gun to a TV screen in 46.0 ns, what voltage was used to accelerate it? (Note that the voltage you obtain here is lower than actually used in TVs to avoid the necessity of relativistic corrections.)
Homework Equations...
Here is a homework question that I have for my general physics class; our professor ran out of time during lecture and did not get to explain the material very well so I am a little lost.
A 47.0 g ball of copper has a net charge of 1.7 µC. What fraction of the copper's electrons have been...
Okay, getting the hang of this.
Last question:
A baseball player initially running at 3.4m/s slides to a stop at third base in 1.2 seconds. Calculate the μk between him and the ground.
Attempt:
I'm not sure how to approach this problem. 3.4m/s divided by the 1.2s will give me the...
Okay thanks for the help, I am feeling kinda dumb now that I know it is that simple.
Next question:
A 60kg snowboarder accelerates down a 32 degree slope at 3.0m/s2. Calculate μk.
Attempt:
Obviously since there is an acceleration, I know that the forces in the x direction acting on...