Asked by AEC
Compare your weight with the attraction (
or repulsion) force of two One Coulomb charges separated one kilometer. Is it larger, smaller or similar?
or repulsion) force of two One Coulomb charges separated one kilometer. Is it larger, smaller or similar?
Answers
Answered by
Damon
I already took the subject. It is your turn. We are here to help, not to do all your work for you. A couple of your questions are interesting but most are standard.
Answered by
AEC
I already did them, just looking to check if I did them alright.
Answered by
Damon
Hard for me to tell.
Answered by
AEC
i.imgur(dot)com/WBQnrHY(dot)jpg
Can you tell me if it is alright?
I'm a bit confused about the rise time and the fall time, I don't know if I should consider fall time at all.
Can you tell me if it is alright?
I'm a bit confused about the rise time and the fall time, I don't know if I should consider fall time at all.
Answered by
Damon
I can see it and am looking
I am wondering if the voltage and current should be linear in time upward during the rise and then linear during fall.
In the last one not only are you neutral charge overall but your individual parts tend to be neutral atom by atom. You do not have a concentration of ions in your feet and - ions in your head :)
I am wondering if the voltage and current should be linear in time upward during the rise and then linear during fall.
In the last one not only are you neutral charge overall but your individual parts tend to be neutral atom by atom. You do not have a concentration of ions in your feet and - ions in your head :)
Answered by
Damon
1 Coulomb charges 1 km apart
E = k Q/r^2
r = 10^3
E = 9*10^9 /10^6 = 9000
EQ = 9000 N
E = k Q/r^2
r = 10^3
E = 9*10^9 /10^6 = 9000
EQ = 9000 N
Answered by
AEC
So do u mean my answer in the first question is wrong?
Answered by
Damon
we need to find the total charge passed during the rise and fall
during rise
i = {10^4amps /[10^-5 s]}t
area under that line (the triangle) is in coulombs
Q = .5*10^4*10^-5 = .05 Coulombs
during fall
Q = .5*10^4 *10^-4 = .5 Coulombs
peak power = volts * peak amps
10^7 * 10^4 = 10^11 watts (yikes)
total energy = integral of Vi dt assuming V is constant that is V int i dt
but we know integral i dt = .05 + .5 =
.55
so 10^7 Volts * .55 coulombs total energy
during rise
i = {10^4amps /[10^-5 s]}t
area under that line (the triangle) is in coulombs
Q = .5*10^4*10^-5 = .05 Coulombs
during fall
Q = .5*10^4 *10^-4 = .5 Coulombs
peak power = volts * peak amps
10^7 * 10^4 = 10^11 watts (yikes)
total energy = integral of Vi dt assuming V is constant that is V int i dt
but we know integral i dt = .05 + .5 =
.55
so 10^7 Volts * .55 coulombs total energy
Answered by
AEC
I don't see where the .5^4 comes.
Thanks for the answer but if you could elaborate it would be awesome.
Thnk so much really.
Thanks for the answer but if you could elaborate it would be awesome.
Thnk so much really.
Answered by
Damon
You mean .5 * 10^4 ?
I am saying that current is proportional to time during the rise (and fall)
It goes from 0 to 10^4 in 10^-5 seconds
the area under that straight line is
.5 * 10^4 *10^-5 which is the charge passed in that time
Therefore my answers should be half yours because I do not us the peak but allow for rise time and fall time
I am saying that current is proportional to time during the rise (and fall)
It goes from 0 to 10^4 in 10^-5 seconds
the area under that straight line is
.5 * 10^4 *10^-5 which is the charge passed in that time
Therefore my answers should be half yours because I do not us the peak but allow for rise time and fall time
Answered by
Damon
remember integral a t dt from t = 0 to t = T = (1/2)a T^2
That is where the 1/2 comes from, area under straight line (triangle)
That is where the 1/2 comes from, area under straight line (triangle)
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.