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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 )

-ost

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

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

4. 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.
Last edited by Psychotron; 09-17-2004 at 02:21 PM.

5. 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.

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

7. 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

8. Originally Posted by 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 = 3.8231E-6, m = 3.8231E-7 kg

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

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

9. 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
Last edited by OSTIE; 09-17-2004 at 01:44 PM.

10. 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.
Last edited by Psychotron; 09-17-2004 at 02:24 PM.

11. Originally Posted by 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.

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.

12. both of our ways are valid for finding the mass, its not possible for it to be incorrect.
Last edited by Psychotron; 09-17-2004 at 03:39 PM.

13. 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

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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?

15. Originally Posted by 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

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

16. Originally Posted by 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?
ugh, dont look at me.

17. Originally Posted by 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?

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