charge of 21micro coloumbs placed in a sphere radius 0.7 meters. what is the flux density through the sphere?
606164
E = k*q/r^2\n\nk = 9 * 10^9\n \nq = 21 * 10^-6\n\nr = .7
725269
Consider the frustum of the cone a surface point of the sphere, then  it's height is given by 0<h<2R (R the radius of the sphere), and because it is inscribed, the base is a planar cut of the sphere, so its radius is given by Sqrt[h(2R-h)] and so the volume of the cone is 1/3*pi*(h(2R-h))h, or VC=pi/3*(2Rh^2-h^3).  Taking the derivative and setting equal to 0, we get 4Rh-3h^2 = 0, or h=4R/3.  The Volume is then (plugging in the formula), VC=32piR^3/81, or 8/27 = (2^3)/(3^3) of the volume of the sphere.\n\nSteve
1054117
from calculus\ndivide the sphere up into circular slices of width dx.\nthe disk at a distance of x from the origin has a radius of sqrt(r^2-x^2) and so an area of pi(r^2-x^2) and a volume of pi(r^2-x^2)dx\nsumming up the volumes of all the disks gives the volume of the whole sphere = int pi(r^2-x^2)dx from x=-r to r\n=pi.r^2.x-pi/3.x^3 from x=-r to r\n=(pi.r^3-pi/3.r^3)-(-pi.r^3--pi/3.r^3)\n=4/3.pi.r^3
895868
Oh dear.  You should be able to do this one.  You simply need to calculate the mass of the core and divide by the total mass.  You have a density and a radius.\n\nMultiply the density of the core by the volume of the core (be sure you use the same units!)  I will leave it to you to look up/remember the equation for the volume of a sphere.\n\nTake the result and divide by the total mass. Are you sure that is the density, btw?  It seems awfully light for iron at the center of the Earth.  Even water has a mass of 1000 kg/m^3.
52468
it is the charge of an electron 1.602*10^-19 Coulombs, i think that is what you are referring to.
625468
The volume of a sphere is:\nVolume =4/3 pi radius^3 \n\nThe radius of the moon is around\n1740 km\n\nSo the volume of the moon would be:\n= 4/3 pi 1740^3\n= 22,066,647,300\n\nSo the moon is about:\n22 billion cu. kilometers, or\n22 quintillion cu. meters.\n\nDon't forget that volume is specified in cubic units.
744501
C = Q/V = εA/d\n\nwhere ε is the permittivity of free space or air\n= 8.8541878176 × 10−^12 F/m (C2/Jm)\n\nQ = (εAV)/d = [(8.85×10^−12)(0.4)(200)]/(0.0005)\n= 1.416 x 10^-6 Coulomb\n\nis the charge on the upper plate.\n\nThe charge on the lower plate that is grounded is ZERO, because the charges have moved down into the ground.
741784
(21 g x 52.7 C) + (54.9 g x 31.5 C) / (21 g + 54.9 g)\n\n= 37.4 C
174332
The surface area of a sphere is A=4*pi*r^2 where r is the radius of the sphere. The *volume* is (4/3)*pi*r^3.
905553
Current = Charge per unit time = 14/7 Ampere = 2 A
981952
For a sphere of constant density having mass M and radius R, the moment of inertia is\nI = (2/5) M R^2\nThe kinetic energy of a travelling and spinning sphere of uniform density is\nK = (1/2) I w^2 + (1/2) M v^2\nK = (1/2)(2/5) M R^2 w^2 + (1/2) M v^2\nK = (1/5) M R^2 w^2 + (1/2) M v^2\nWhere w is the angular speed.  And where v is the speed of the sphere's center of mass with respect to the surface on which the sphere is moving, which is equal to the tangential speed of the sphere's rotational "equator," which is in contact with the surface on which the sphere is moving, with respect to the sphere's center of mass.\nw = v / R\nK = (1/5) M R^2 w^2 + (1/2) M v^2\nK = (1/5) M v^2 + (1/2) M v^2\nK = 0.7 M v^2\nK = 0.7 M R^2 w^2\nFor your sphere, M=0.5 kg and R=0.06 m, so...\nK = (0.00126 kg m^2) w^2\nOn a slope 30 degrees from horizontal, the acceleration from gravity would be\ng = (9.8 m/s^2) cos(30) = 4.9 m/s^2\nThe work done by a ball rolling one meter along the slope is\nE = M g (1 meter) = 2.45 Joules\nK = E\nThe angular speed of the sphere after rolling 1 meter is\nw = { K / (0.7 M R^2) }^(1/2)\nw = { 2.45 Joules / (0.00126 kg m^2) }^(1/2)\nw = 44.096 s^-1\nv = wR = 2.6458 m/s\nThe average acceleration of the ball is found from the regular kinematic formula.\nVf^2 - V0^2 = 2 a dx\nWhere dx is the one meter measured along the slope, V0=0, Vf=v.  The average acceleration, therefore, is\na = v^2/(2 meters) = 3.5 m/s^2\nSo the speed of a sphere of constant density having mass 0.5 kg and radius 0.06 meters, rolling freely down a 30 degree frictionless slope, is\nv = (3.5 m/s^2) t
558215
F = KM1M2/r^2\n\nwhere K is the constant, M1 and M2 are the magnetic moments, and r is the distance between them.
1146296
The magnetic flux:\nit is the product of area & magneric flux component \nThe magnetic flux density:\nit is numerically measured by the number of magnetic flux lines passing perpendicular to a unit area surrounding that point & it is measured in Tesla
801826
E=hf  where h = 6.6210^-34 Js\n\ne = 6.6210^-34*4.31*10^14 = 2.71 10^-19 Joule=1.7 eV
101953
There's a simple way to go about this. The answer is reasonably accurate too.\n\nNow, we know that the formula for the volume of a sphere is 4/3 times pi times the cube of earth's radius.\n\nGiven the value of pi as 3.14 and the earth's radius as 6400 km, and also assuming that the earth is a perfect sphere, we get the volume as 1.09806622 * 10^12 cubic kilometers.
125839
I'm not sure. But the volume of a sphere is (4/3)(pi)(r)^(3). \n2) You know that the diameter is double the radius. So your radius is 1.\n3) Then, your volume is (4/3)(pi). \n4)Density=(mass/Volume)\n5)Density of water is 1.00 mL=1.00g/m^3\n6)MASS=(Density)(Volume)=((1.0g/m^3)(4/3)pi). Note that cubic meters cancel. The mass is (4/3)pi grams.\n\nHope I helped.
91154
If there are charges on the sphere, then the moving charges will produce a magnetic field. This is called the Biot-Savart law.