A charged particle A exerts a force of 2.29 μN to the right on charged particle B when the particles are 13.5 mm apart. Particle B moves straight away from A to make the distance between them 17.8 mm. What vector force does particle B then exert on A?
I got 1.317 E-6, but apparently that's not right. ???
do your own homework
Physics will get you nowhere.
uh_no wrote:
do your own homework
I am. I'm pretty sure I'm right and the computer is wrong, which is why I came here to see. I solved it two different ways and I got the same answer both times, soooo yeah.
Ask Garret Health....PHD student in physics
Not going to do the math, but the quantity would be negative. The vector force on B from A is positive, the vector force on A from B must be negative.
Throw physic to the dogs, I'll none of it!
P.Whelan wrote:
Not going to do the math, but the quantity would be negative. The vector force on B from A is positive, the vector force on A from B must be negative.
I don't need the positive or negative, just the magnitude.
Bumping for the OP. Cmon smartasses, someone help this dude.
that is a correct answer with inverse-square relationship
ventolin^3 wrote:
that is a correct answer with inverse-square relationship
On another problem I had to find acceleration, distance, and time, with a known velocity. My answers to acceleration and distance were right, but it says my answer for time is wrong even though the equation used is simply v=at. I also used x=.5at^2 and got the same answer, so I'm just getting kind of pissed off at the program.
ventolin^3 wrote:
that is a correct answer with inverse-square relationship
Yep, I got the same thing.
I think I'm right wrote:On another problem I had to find acceleration, distance, and time, with a known velocity. My answers to acceleration and distance were right, but it says my answer for time is wrong even though the equation used is simply v=at. I also used x=.5at^2 and got the same answer, so I'm just getting kind of pissed off at the program.
makes no sense apart from there usually being a "u" in
v = at
& u*t in
s = 0.5a*t^2
this is all primeval...
I think I'm right wrote:
A charged particle A exerts a force of 2.29 μN to the right on charged particle B when the particles are 13.5 mm apart. Particle B moves straight away from A to make the distance between them 17.8 mm. What vector force does particle B then exert on A?
I got 1.317 E-6, but apparently that's not right. ???
Your answer is wrong because you have not included your units (not to mention direction, since this is a vector quantity, but you've addressed that in another response).
If I were a computer program, I would expect the answer to be expressed in the same units that were given in the problem. In which case, you would again be wrong, because the correct answer is not 1.317e-6 μN.
a physics teacher wrote:
Your answer is wrong because you have not included your units (not to mention direction, since this is a vector quantity, but you've addressed that in another response).
If I were a computer program, I would expect the answer to be expressed in the same units that were given in the problem. In which case, you would again be wrong, because the correct answer is not 1.317e-6 μN.
This is why I despise those computerized homework things (Mastering Physics is what they used when I TA'd). They pointlessly frustrate students who are struggling to understand the subject in the first place.
-A physicist.
E = MC^2.
I understand the physics; I got a 5 on the AP exam 4 years ago in high school. For some reason my school only gave me credit for 1st semester physics, even though I was taught what we're doing in physics II as well.
Anyway, the computer program didn't need the negative sign since it was stated in the question. The problem was that it wanted it in micro N instead of just N. I'm used to always putting my answer in standard SI units; no milli, micro, nano, etc, so that's where the confusion was.
Thanks for the help though, guys.