Thanks!
Ok got it...took a moment to figure out the addition of the fractions in parallel...but sure enough it worked :)
I think the V^2 / R = P was the part that cracked it for me. P = IV wasn't seeming to swing it for me :)
In case anyone else needs it and for someone else to double...
Homework Statement
The circuit in the drawing contains five identical resistors. The V = 50 V battery delivers 46 W of power to the circuit. (Picture should show up somewhere)
Q: what is the resistance in each resistor
Homework Equations
so far the only things i have...
9 in the denominator is from 3^2 as r^2
the only other thing i have come up with is since it is the repulsive force...then 3 sqrt(2) so 18 in the denominator giving me .405
.405 and .72 are my two sides then?
God i don't get this problem...
i got all the starred, double starred, etc. problems...and this no star basic arse problem i just am not getting. Why on Earth is this one so hard for me LOL
F***
bleh. sorry the problems are getting jumbled in my head...
F = 9e9 * 45e-6 * 18e-6 / 9 = .81 N
and we already determined .72 N
So, now i calculate the x and y component?
i get: x1 = 0, y1 = .81
x2 = .72, y2 = 0
pyth. theo. => (.81^2+.72^2)^.5 = 1.08N?
ok...it asks for: kq / x^2+y^2 * cos @ (where @ = theta).
only thing i can think of is 45 degrees...so:
kq / x^2 + y^2 * cos (45) = ((9e9 * 16e-6) / (3^2))* cos (45) + ((9e9 * 18e-6) / (3^2))* cos (45)?
(currently working on this...i realize i forgot the x and y component part of the...
It is two parts:
https://www.physicsforums.com/showthread.php?t=124995
this thread has the pictures.
I understand it is E = kq / r^2...
a) i pythagorean theoremed the hypoteneuse... so (1.5^2 + 2.25^2)^.5 = 2.704 cm
2.704 / 100 = .02704 m
8.99e9 7.5e-12 / (.02704^2) = 92.31 N/C...
Ok so i was just not htinking in vectors...but here is what i have (i only have one more chance at this before it is due so i am trying to be cautious).
K * 18e-6 * 16e-6 / 9 = .288 = y comp1
K* 18e-6*45e-6/9 = .81 = x comp1
K*45e-6*18e-6/9 = .81 = y comp 2
K*45e-6*16e-6/9 = .72 = x...
Ah ha! yes apparently the one thing I didn't do was check for stupid mistakes on this one :)
Any chance you can give me some guidance on the 2nd part? Thanks!
If they are straight on the x-axis just out 3.0m and the other one straight up on the y-axis just up 3.0 m...then doesn't that make them 90 degree angles? meaning no x component on the charge on the y-axis and then for the charge on the x-axis there is no y component?
So, just a charge 3.0 m...
I saw that and read it and hten posted this one. Based on what i picked up in that thread, I ended up with the above.
However, based on what you said I assume if we drew it out on a line...with the negative charge as our starting point...2.32m out would be what I said was the answer...
Homework Statement
The drawing shows three point charges fixed in place. The charge at the coordinate origin has a value of q1 = +9.00 µC; the other two have identical magnitudes, but opposite signs: q2 = -5.00 µC and q3 = +5.00 µC.
For the problem imagine the x-y coordinate grid with a +...
Homework Statement
Two charges, -26 µC and +5 µC, are fixed in place and separated by 1.3 m.
(a) At what spot along a line through the charges is the net electric field zero? Locate this spot relative to the positive charge. (Hint: The spot does not necessarily lie between the two charges.)...
Homework Statement
#1:
Four point charges have the same magnitude of 6.80 x 10^-12 C and are fixed to the corners of a square that is 1.0 cm on a side. Three of the charges are positive and one is negative. Determine the magnitude of the net electric field that exists at the center of the...
Homework Statement
Three charges are fixed to an x, y coordinate system. A charge of +18 µC is on the y-axis at y = +3.0 m. A charge of -16 µC is at the origin. Lastly, a charge of +45 µC is on the x-axis at x = +3.0 m. Determine the magnitude and direction of the net electrostatic force on...