A point charge q is placed distance r from the centre of an uncharged conducting sphere of radius R. What’s the potential at a point on the surface?
An aeroplane has to go from point A to another point B, 500 km away due 30 degrees east of north. A wind is blowing due north at a speed of 20 m/s. The air-speed of the plane is 150 m/s. What direction should the pilot head the plane to reach point B?
The vector diagram for various velocities is shown.Vector AC gives velocity of aero plane relative to quiet air.Vector CB gives velocity of wind(air) relative to ground.Vector AB gives velocity of plane relative to ground.Using law of sines for triangle,(150/sin 30 degree) = (20/sin theta).Therefore, sin( theta)=(20/300)=(1/15). Then,Theta=sin^-1(1/15)=3.8 degree.Angle NAC=33.8 degree.The pilot should head the plane in the direction 33.8 degree east of north.
How can you calculate the Coulomb force between the proton and electron of a hydrogen atom?
You can calculate the Coloumb force between any two charged bodies using Coulomb’s Law:F = k (q1*q2/r^2)Where k is Coloumb’s constant:8.99*10^9Nm^2C^-2The electron carries the elementary charge: q = -e.The proton carries the same charge but positive q = eThe value of e is empirically measured to be: 1.602*10^-19 CNow, the r value is a little tricky. r is meant to be the distance between charges. Unfortunately, thanks to quantum physics, the electron could be any distance from the proton. It only has a certain probability of being a given distance.Luckily, physicists account for this by giving a value called the Bohr radius. This is the average or most likely distance a single electron would be from the centre of the single proton in a hydrogen atom. This value is defined as: 5.291772*10^-11mSo we have all our values! Let's plug ’em in and see what we get.F = k (q1*q2/r^2)F = (8.99*10^9Nm^2C^-2)(((-1.602*10^-19 C)(1.602*10^-19 C))/(5.291772*10^-11m)^2)F = (8.99*10^9)(-9.141*10^-18)F = -8.22*10^-8NNote that the value is negative because the proton and electron are attracted to each other. As well, the value would vary depending on where the electron was at any given moment and so this is only the most probable value.
Why is gravity zero at the centre of earth?
Yes, the gravitational force at the earth’s center is zero.Here is the explanation:-The further you go away from earth, lesser the gravitational force you experience. And closer you come to the surface of earth, more gravitational force acts on you. This is because of the inverse square law ( See:- Newton's law of universal gravitation - Wikipedia ).But as you keep on moving towards the center of the earth, the gravitational force acting, by the virtue of earth’s mass, on you will gradually abate and ultimately go zero once you reach the earth’s center. Not because there is no gravitational force down there but because at the center you will be completely surrounded or enveloped by earth, as a result you will be pulled by equal forces in all directions. Hence all the gravitational force lines will cancel out each other leaving you with zero net force.So apparently you won’t feel any gravitational force down there at the center.Image Credits:- Google
Math help calculus III?
cos(-5 + 5) = 1 = e^(-5 * 2 + 10); so (-5, 5, 2) is on the surface. (a) Let f(x,y,z) = e^(xz + 10) - cos(x+y). So, computing the gradient for a normal vector to the surface yields ∇f =
Area of triangle on the argand plane formed by complex numbers z,iz, z+iz is?
The answer is |z|^2/2let A= z, B=iz and C= z+izlet z=x+iy then if you take out the value of |A-B|, |B-C| and |C-A| you will see that it is a n isosceles triangle with AC=BC.It is a property of an isosceles triangle that if we join C and mid point of A and B (let mid point be P), it will be perpendicular to AB.AREA = 1/2 *AB*PCP= mid point of A and B = A+B/2= (z+iz)/2now PC= |z+iz-(z+iz)/2|= |(z+iz)/2|AB= |z-iz|area= 1/2*|z+iz|/2*|z-iz|= |z^2+z^2|/4=|z|^2/2