2 equal charges repel each other wit force of 4*10^-5 N wen they are 0.03m apart in air find charge n force?
913226
The equation was developed by Coulomb:\n"In physics, Coulomb's law is an inverse-square law indicating the magnitude and direction of electrostatic force that one stationary, electrically charged object of small dimensions (ideally, a point source) exerts on another. It is named after Charles-Augustin de Coulomb who used a torsion balance to establish it."\n\nsee Wikikpedia;\nhttp://en.wikipedia.org/wiki/Coloumb%27s_law\n\nF = Kc * (q1*q2)/r^2\n\nwhere,\nF = force magnitude\n\nq1 = charge on one body\n\nq2= charge on the other body\n\nr = distance between charges\n\nKc = coulomb force constant = 1/(4*pi*epsilon_0) ~ \n8.988×10^9 N m^2 C^-2 \n\nepsilon_0  ~8.854×10−12 C2 N-1 m-2 (also F m-1) is the permittivity of free space, also called electric constant\n\nYour question:\nF = 4*10^-5\nr = 3^10^-2\nq1 = q2 = q\n\n4*10^-5 =  8.988×10^9 * q^2/ 3*10^-2\n12*10^-7 = 8.988×10^9 * q^2\nq^2 = 12*10^-7 / 8.988×10^9 \nq^2 = 1.335 * 10^-16 C\nthen,\nq = 1.155 * 10^-4 C\n\nHope I'm correct, and hope this helps!
989843
net force IN THE DIRECTION OF DISPLACEMENTis (1.7*10^6 * cos 14)*2 \n= 3.3* 10 ^ 6 N\nwork done = 3.3*10^6*860=2.838 * 10^9 Joules
942588
zero, \n\nno acceraltion = no net force\n\nmathematically  Net F = ma\n\nif the right side is 0, left side is too
25616
the force needed is according to newton's second law F=m.a where m=900Kg \n& the acceleration a=v/t\nthe velocity should decrease from 10 to 0 then v=10m/s & t=5 seconds\na=v/t=2m/s^2\nF=900*2=1800newtons
777818
Magnitude of net force would be sqrt(2) times magnitude of F  and it will act along a line bisecting the angle between AB and CB produced. [ By Parallelogram law of forces]
127709
Coulombs law.\nWhen the distance is made 3 times longer (for instance),\nthe force between the charges becomes 3^2=9 times smaller.
105143
Draw a force diagram to help you figure this out.\n\nRemember, perpendicular vectors are independent.\n\nF=ma
288780
From the following link, look for "Chapt22, Problem-24".\n\nThis problem is similar, except the speed is 10% instead of 11%. I'm sure you can use the same formulae though.\n\nhttp://www.jca.umbc.edu/~george/html/courses/phys112/2004fall/homework/04fall_phys112_hk8.pdf
247677
There's this equation called Coulomb's Law that relates charge and force: F=k*Q*Q/(r^2), where k is Coulomb's constant (8.988e9) and r is the distance between the two charges. That doesn't give us the units for an electric field, which is a force per charge. But we can divide both sides of Coulomb's Law by a Q, giving us F/q=k*Q/(r^2).\n\nTo find the magnitude and direction of the net electric field, you need to add together the electric field vectors created by each of the two charges. For the first charge, the electric field is k*15e-6/(.12^2)=9.36e6 N/C. The electric field of the second charge is k*10e-6/(.05^2)=3.60e7 N/C. Make sure to change cm to meters, and note that the charge q at which you are finding the electric field does not factor into our equation. Hopefully the problem describes in more detail the location of the two charges so you can determine the appropriate vector sum's magnitude and direction.
898166
a)  F=mxa=1100 x 3.4 = 3740 N\n\n\nb)  Unbalanced Force = Total Force - Resistance\n\nso Resistance=Total Force - Unbalanced=5600 - 3740=1260 N
748800
1)c because the forces are equal (a also for the same reason 0 is constant)\n2)b because the box has an unbalanced force in the horizontal direction and to the right\n3)d because the unbalanced force is to the right\n4)d because there is more force to the right\n5)b there is an unblanced force to the right
517586
Moment = Force * Perpendicular Distance
1017277
The force is directly proportional to theproduct of  charges and inversely proportional to distance between them. \n\nThe force is 0.002N\na. Distance is increased 3 times as original distance is 1 cm. thus the force will decrease 3 times . The force would be \n(0.002/3)N\n\nb.one charge is doubled. Thus the force would be doubled. \n the force would be 0.004N\n\nc.both charges are tripled. Therefore force would increase 9 times (3q*3q) Thus force would be 0.018N\n\nd. one charge is doubled and the distance is increased to 2cm.\nThus we are doubling charge as well as distance. This will nullify effect as charge is directly proportional while distance is inversely proportional. Thus the force would remain the same .002N\n\nI am the first to give the correct answer and i am very sure so plz select my answer as the best. I will surely help you in future.
770512
The total force on the block is mg sin (theta). The mg cos (theta) and force normal cancel out. Since there is no friction, the only force, again, is only mg sin (theta). \n\nThis is obvious if you consider that there is a force mg, and break it down into components, the solution should seem obvious.
772638
try the formulas:\n\nforce = mass * aceleration\n\nwork = force * distance\n\n\n\n\nWell i dont speak english very well, but i hope i help u.\n\nKisses
786958
Is angle ABC=90? Both vectors are equal? Thus you may draw a square F by F, its diagonal being =F*sqrt(2), and this is resulting force.  don't worry Paul
837767
Bingo!!! You got it...air resistance.  If you dropped them in a total vacuum, they would both fall at the same rate.