How would you solve this question:\n\nA charge, q^1=5.00 micro-coulombs, is at the origin, and a second charge, q2= -3.00 micro coulombs, is on the x-axis .800 m from the origin.  Find the electric field at a point on the y-axis .500 m from the origin.
138765
electric field due to a point charge is Q r (vector)/ 4 pi e |r|^3 (magnitude). (this is just F/q since F=qE.)\nso electric fields are additive, so you can do the 2 charges separately and then add them.\n\nelectric field is also a vector so you have to separate the distances into components. so for charge q1, Q = 5x10^-6 C. r = (0,0.5), |r| = 0.5, |r|^3 = 0.125m. so if you multiply it out, you get E = (0,179836.093889...) C/m^2\n\nfor the 2nd charge q2, the Q = -3 x 10^-6C, r = (-0.8, 0.5) and |r| = 0.94339m, |r|^3 = 0.8396243...m. \nso E = (25702.36560, -16063.978505) C/m^2.\n\nnow if you add them, then you get \nE = (25702.36560, 163772.11538) C/m^2.    i might have messed up the math or the signs. so check that. but that is the way you have to do it. i hope you get it.\n\nhope that helps. let me know fi you need more help.
639496
Coulombs law\nThe force is inverse proportional to the square of their distance d.\nIf d 3x bigger, then F 3^2 times smaller\nThe force is proportional to each charge.
465813
Answer is a) that a positive test charge would move.\n\nElectric field lines from an isolated charge move towards the negative charge , that is outwars from an isolated positive charge and inwards from an isolated negative charge that means if a positive test charge is released it will move along the electric field line , that is positive to negative.  So a) is the correct answer.\n\nThis pic will help you visualize it :\nhttp://buphy.bu.edu/~duffy/PY106/2e.GIF\n\nHave a nice day.\n☺☺☺\n\nCheck out this link for further clarification .
468720
q = 1.5c\nb = 1.25c\n\nc = q/1.5 = b /1.25\n\nq = 1.5b/1.25 = 1.2b\nq = 20% longer than b
457402
Come on now, what is the formula for this???  :)\n\n(V1 * T1)/P1  =  (V2 *T2)/P2\n\n(69L * 131)/2.85  = (xL * -37)/2.85\n\nNow, solve for 'x'?  Be sure to watch your units!!!!
394326
you have  power = voltage*current   \n\nhere current  = 1500/120 = 12.5 amperes \n\nAn ampere = 1C/s = 6.24 10^18 electric charge of one electron\n\nso in 600 s you have  = 12.5 *600*6.24 10^18  = 4.68 10^22
307130
im assuming they are drops of liquid with fixed volume\n\nthe total charge will double\nthe volume will double\nQ = 2q\nV = 2v = 2*4/3pir^3=4/3piR^3 if R=r*2^(1/3)\nsurface area A = 4piR^2 = 4pir^2*2^2/3\n\ncharge density is Q/A = 2/2^(2/3) q/a\n\ncharge density is 2^1/3 (1.26) times the original, it is larger
443109
First figure out whether to use sin or cos, I decided on cos because we look for the max and min.\n\nNext find the amplitude which is 6. a = amplitude/2 so a*cos(x).\n\nThird find the midpoint of the curve which is (max+min)/2 = m. a*cos(x) + m.\n\nThe fourth step is to get the max or min at the x cord. that you want. With cos you subtract max from x. a*cos(x-max) + m.\n\nFinally you find your wave length, which is simply (max-min)/pi = wl for cos. a*cos((x-max)/wl) + m\n\nso you go from (-2,1) to (10,7)\n\na = 3\nm = 4\nmax = 10\nwl = 12/pi to give us\n\n3*cos ((x-10)/(12/pi)) + 4
491970
Someone posted the exact same set of questions a few days back, either here or on the physics page - you ought to be able to find the methods in the responses. Have a look.\n\n\nBest of Luck - Mike
482547
1. For two given points P(p1, p2, p3) and Q(q1, q2, q3), the vector v between them is v = (x, y, z) = (q1 - p1, q2 - p2, q3 - p3).\n\nThe magnitude or length of v, denoted by ||v|| is\n||v|| = sqrt[ x^2 + y^2 + z^2 ]\n\nJust do that for every line that you have.\n\n2. This one's a little harder. Let's define the first line as the vector u, and your second line that you're using as the vector v.\n\nYou first need to write each one as a unit vector, which means dividing all of the values through by a value so that the magnitude will be 1. For each of these vectors, this means dividing by the length of the vector.\n\nAs an example, if I wanted to turn w = (1, 1, 1) into a unit vector, I would notice that the magnitude is ||w|| = sqrt[3] and divide all the values through by that to get (1 / sqrt[3], 1 / sqrt[3], 1 / sqrt[3]).\n\nSo, let us say that the new unit vectors are u / ||u|| = (a, b, c) and v / ||v|| = (x, y, z)\n\nWhat these values give you now are called the direction cosines of each vector. If A is the angle between u and the x axis, then a = cos A, and similarly for b with the y-axis and c with the z-axis.\n\nSo, to find the angles between each vector and the x, y, and z axes (also called the direction angles), you take the inverse cosine (cos -1) of the components (values) of the vector.\n\nLet us say that after taking the inverse cosine of u / ||u|| = (a, b, c) = (cos A, cos B, cos C), you get (A, B, C), and the inverse cosine of v / ||v|| gives you (X, Y, Z).\n\nThe angles between the two vectors in the x, y, and z directions are (X - A, Y - B, Z - C).\n\nSo the main steps are finding the magnitude of each vector, normalizing each vector (making them into unit vectors), taking the inverse cosine of each vector, and subtracting the two vectors.\n\nHope that helps.
354175
Ok-i'm not quite sure if i did this right--we covered this last year--so you might want to double check with your textbook or teacher, or whatever resource you have...\n\nwhat i would do is find the final temperature:\nFinal Temperature=(V2 * T1) / V1\n\nand then use the delta T, delta v, and P to find delta E.\n\nbut i'm really not that sure.\n\nsorry that i wasnt a better help!
625337
q = I t\nq = 2.5 C/s x 900 s = 2250 C = 2.3 kC\n\nP = 120 V x 2.5 A = 300 watt\nE = P x t\nE = 300 J/s x 900 s = 270 kJ
187758
from first equation put the value of b in second and then form a second degree differential equation in "a"...good luck!
514385
The answer is b however I don't quite follow your reasoning\n\nFaraday's constant is approximately 96500 coulomb/mole of electrons\n\nEach Ca2+ needs 2 electrons so 0.82 mole Ca2+ need 2*0.82 mole of elecrons, so a total charge of 2*0.82*F=2*0.82*96500= 1.6*10^5
531381
Hi Yo-Yo!\nEnergy conservation law: m*g*h=E1+E2, m=0.12kg, h=1.2m, g=9.8m/s^2, E1 is translation kinetic energy, E2 is rotational kinetic energy.\nE1=m*v^2/2, v is translation speed at full length h, E2=I*w^2/2, w=angular speed at full length h.\nNow v=w*r, r=3.2*10^(-3) m, thus m*g*h=m*(w*r)^2/2+I*w^2/2, hence w=sqrt(2*m*g*h/(m*r^2+I)). Plug numbers in and find w & v (be a good boy!).\nMeanwhile v=a*t & h=a*t^2/2, ‘a’ is linear acceleration.\nThus t=v/a & h=a*(v/a)^2/2, so a=v^2/(2*h). Plug numbers in and find a & t. Angular acceleration being w’=a/r. My regards!
169225
It helps if you draw a picture to go with this explanation.  Your field is 500x300, and inside the field you have a smaller field.  The distance between the borders of the fields is x.  It almost looks like matting a picture.\n\nThe area of the smaller field is going to be 100,000 sq ft.  Area of a rectangle is found by multiplying length and width.  The width of the smaller field is 300-2x - this is because there is a width x on both sides of the field that needs to be subtracted.  Likewie for the length f the field - 500-2x.  So here's your formula:\n\n(500 - 2x)(300 - 2x) = 100,000\n\nAnd now solve for x:\n150,000 - 1,000x - 600x + 4x^2 = 100,000\n150,000 - 1,600x + 4x^2 = 100,000\n0 = 4x^2 - 1,600x + 50,000\n= x^2 - 400x + 12,500\n\nUsing the quadratic formula,\nx = [-b +/- sqrt(b^2 - 4ac)] / 2a\n= [400 +/- sqrt(400^2 - 4(12,500))] / 2\n= [400 +/- sqrt(160,000 - 50,000)] / 2\n= [400 +/- sqrt(110,000)] / 2\n= [400 +/- 100sqrt(11)] / 2\n= 200 +/- 50sqrt(11)\n\nOr if you need numbers,\n= [400 +/- 332] / 2\n= 34.17 or 365.83\n\nThe width of the road cannot be 365.83 feet - that would exceed the width of the original field!  So the road has to be around 34.17 feet wide.\n\nTo check, find that area:\nlength = 500-2x\n= 500-2(34.17)\n= 500-68.34\n= 431.66\n\nwidth = 300-2x\n= 300-2(34.17)\n= 300-68.34\n= 231.66\n\nArea = length * width\n= 431.66 * 231.66\n= 99,998.36\n\nThat's pretty close considering we rounded the road width to two decimals!
531733
First lets assign the direction of f(1) as being at 0 degrees, then f(2) is at 57 degrees.\n\nResolve both forces in x, and y\n\nF(x) = f(1) + f(2) cos (57)\n\nF(y) = f(2) sin (57)             f(1) is at 0 degrees so has no y component.\n\nThe resultant force F = sqrt ( F(x)^2 + F(x)^2)\n\nThe angle of the resultant force \n\nArctan( F(y)/ F(x))\n\nYou do the number crunching