PLEASE HELP!! w/ physics problem...

[OSTIE]

Alright, I know there are some smart fckers on here. I need yea guys to lend me a hand here. I have some stupid physics assignment due this evening, but I havent bought the book yet and Im stumped on the last two problems. I think i can get one of them, but here is the other.

Two equally charged particles, held 3.1 10-3 m apart, are released from rest. The initial acceleration of the first particle is observed to be 8.0 m/s2 and that of the second to be 10.0 m/s2. The mass of the first particle is 6.3 10-7 kg.

(a) What is the mass of the second particle? (in kg)

(b) What is the magnitude of the charge of each particle? ( in C )


Thanks in advance,

-ost

[Psychotron]

id help but im unclear about your 3.1 10-3m and the 6.3 10-7kg, what the heck is that.

[PT]

open up your physics book and find the answer. i took 1 class as a elective and hated it.

[Psychotron]

F = k(q^2)/r , F = m*a , q^2 since they are both equally charged

8.0(6.3E-7)=k(q^2)/3.1E-3 where k = 8.99E9

5.04E-6=8.99E9(q^2)/3.1E-3
1.5624E-8=8.99E9(q^2), q^2 = 1.73793E-18, q= 1.31831E-9

b) q = 1.31831 x10^-9 C

a) 10(m) = 8.99E9(1.31831E-18)/3.1E-3 , 10m = 5.04E-6, m = 5.04E-8 kg

i think thats right, if i remember coloumbs law correctly.

[Natural Mac]

A. F1 = F2... M1*A1 = M2*A2 ... So Simple!!!

B. Use Coulombs Law, q1 = q2. Just solve for q.

You really need to try harder. This is first year crap! Have fun.

[hung-solo]

wtf? e=mc2......maybe? i am glad i stuck with american history and accounting

[OSTIE]

yea psych. those are the correct formulas. Thanks for the help it seems correct. And you guys are correct, it is easier crap that I should know. This whole first two weeks is mostly review from the previous ph. class, but i forgot alot over summer.

I also have not bought the physics book yet for me to read, I have to wait until my next paycheck before I can afford the other half of my books, so until then im stuck.

Thanks for the help,

-ost

[tryingtogetbig]

F = k(q^2)/r , F = m*a , q^2 since they are both equally charged

8.0(6.3E-7)=k(q^2)/3.1E-3 where k = 8.99E9

5.04E-6=8.99E9(q^2)/3.1E-3
1.5624E-8=8.99E9(q^2), q^2 = 1.73793E-18, q= 1.31831E-9

b) q = 1.31831 x10^-9 C

a) 10(m) = 8.99E9(1.31831E-18)/3.1E-3 , 10m = 3.8231E-6, m = 3.8231E-7 kg

i think thats right, if i remember coloumbs law correctly.

dam.....beat me to it......

[OSTIE]

actually, i noticed psyc. formula is wrong.... you didnt square the distance r

F = [K(q1)(q2)/r^2]*[r_vector] I understand that vectors do not matter in this problem, but i just posted the complete formula anyways.

I also did it this way before i originally posted, but I got the wrong answer so thats why I posted. I dont know if im entering it in wrong or what?

BTW, i am using the F = ma, so ma = K*q^2/r^2

I enter the answers on webassign, so i know if they are correct or not right away

[Psychotron]

oh, yea the vectors supposed to be squared. i last took physics a year ago lol

my numbers arent the same as what mac was saying. im not sure who is right. i assumed the coloumbs equation since they give you a vector distance.

so even with the r^2 the answer is wrong?


8.0(6.3E-7)=k(q^2)/(3.1E-3)^2 where k = 8.99E9

5.04E-6=8.99E9(q^2)/(3.1E-3)^2
4.84344E-11=8.99E9(q^2), q^2 = 5.38758E-21, q= 7.34E-11

b) q = 7.34 x10^-11 C

a) 10(m) = 8.99E9(5.38758E-21)/(3.1E-3)^2 , 10m = 5.04E-6, m = 5.04E-8 kg



MACS Way

F1 = F2 , M1*A1 = M2*A2 8(6.3E-7)=10M2 , M2 = 5.041E-8kg

our masses line up.

note: i screwed up in my first time on the mass and later edited it when i saw the mistake when i did the calculations with the vectors squared.

[OSTIE]

oh, yea the vectors supposed to be squared. i last took physics a year ago lol

my numbers arent the same as what mac was saying. im not sure who is right. i assumed the coloumbs equation since they give you a vector distance.

so even with the r^2 the answer is wrong?


8.0(6.3E-7)=k(q^2)/(3.1E-3)^2 where k = 8.99E9

5.04E-6=8.99E9(q^2)/(3.1E-3)^2
4.84344E-11=8.99E9(q^2), q^2 = 5.38758E-21, q= 7.34E-11

b) q = 7.34 x10^-11 C

a) 10(m) = 8.99E9(5.38758E-21)/(3.1E-3)^2 , 10m = 5.04E-6, m = 5.04E-8 kg



MACS Way

F1 = F2 , M1*A1 = M2*A2 8(6.3E-7)=10M2 , M2 = 5.041E-8kg

our masses line up.

note: i screwed up in my first time on the mass and later edited it when i saw the mistake when i did the calculations with the vectors squared.

Finally got the force correct, I had the power one off, i tried your power and it was right. I need a friggin ti-89, im stuck using a plain old calc.

ANYWAYS, the mass one doesnt seem to be correct... webassign though is a pain, because sometimes if you round off to early, you will be a hundreth off and itll be wrong.

[Psychotron]

both of our ways are valid for finding the mass, its not possible for it to be incorrect.

[Psychotron]

well it wont let me edit it so here. i did both way in one step on my ti-89 and the mass for both comes to 4.8E-7 kg

[SAUCYgator]

i respect the question and the fact that you recieved a legitimate answer.... now who wants to take my genetics test for me next week?

[OSTIE]

well it wont let me edit it so here. i did both way in one step on my ti-89 and the mass for both comes to 4.8E-7 kg

5.0e-7 is what the friggin website took as an answer. I hate this friggin software that they use for us to submit answers.

Thanks for the help, it is greatly appreciated,

-ost

[Psychotron]

i respect the question and the fact that you recieved a legitimate answer.... now who wants to take my genetics test for me next week?

ugh, dont look at me.

[tryingtogetbig]

i respect the question and the fact that you recieved a legitimate answer.... now who wants to take my genetics test for me next week?

I'll take it...let me know when and where....I ain't sceered!!!!