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If you are standing with your heels against a wall, you cannot touch your toes.Why?
(Asked Camille )
But when we are standing against a wall, the body cannot adjust itself to adjust the centre of gravity within the base as the backward motion is restricted by the wall. Therefore when we try to lean forward, the CG is shifted out of the base and we tend to fall.
For a circular motion centripetal force is required. Is there any other force which could replace centripetal force to obtain a circular motion? (Asked Sajin)
Please note that Centripetal Force is the net force that must be acting towards the centre of the circular path. It may be provided by one or more of the forces of different kinds. It is not a kind of force like the frictional force or gravitational force. In different cases the centripetal force is provided by different forces.
For example, when we whirl a stone tied to the end of a rope in a horizontal plane, the tension in the string provides the centripetal force. Here the real force acting is the tension in the string and its magnitude equals the centripetal force mv^2/r.
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Two skaters, one with Mass 45kg, 2’nd with Mass of 65kg stand on an ice rink holding a pole with L=10m with negligible mass. starting from the ends of the pole, the skaters pull themselves along the pole until they meet. How far does the 45kg skater move?
Hint: They will meet at the centre of mass of the system.
If the 45 kg skater moves by x m, then the second one travels (10-x)m
45 x = 65(10-x)
Solve and get the value for x
Rajkumar asks: “accelaration due to gravity decrease on depth of earth with the relation of g’=g(1-2h/R). but with relation of g=GM/(R*R) g increase.
When the object is on the surface or above it, the entire mass of the earth attracts it. But when it is inside, the net attraction towards CENTRE is provided only by the mass of earth below it.
ie; if it is at a depth d below the surface, then it is attracted to the centre by a sphere of radius R-d only. The forces due to the rest of the mass of earth cancel out. (On one side there is a little mass and on the other side greater amount of mass but distributed faraway and both contribute equally towards force and cancel out.)
†ђe fig below shows a tilt hammer ahinged at o, with its head resting on top of †ђe pile Bm †ђe hammer including †ђe arm OA, has a mass of 25kg.its centre of gravity Gis 400mm horizontally. from O Ąπϑ its radius of gyration about an axis through G parallel to †ђe axis of †ђe pin 00000 is 75mm. †ђe pile has a mass of 135kg. †ђe hammer is raised through 45• To †ђe position shown in dotted lines Ąπϑ released. On striking †ђe pile there is no rebound. Find †ђe angular velocity of †ђe hammer immediately before †ђe impact Ąπϑ †ђe linear velocity of †ђe pile after impact. Neglect any impulsive resistance offered by †ђe earth into Ψђíçђ †ђe pile is being driven. [Thank you]
- Magnitude and direction of dipole field depends not only on distance r from centre of dipole,but also on the angle between the position vector rand the dipole moment p how?
- How can centres of positive and negative charge lie on same place and if both have same charge they should become neutral?
1. The electric field at a point due to a dipole is given by
Which clearly explains the answer to your question
2. Centre of positive charge means the centre of mass of all the positive charged particles and similarly for the negative charge. If the atom or molecule is symmetric, the two centres of charge will coincide; just like the centre of mass of two concentric hollow spheres. The charges are not at the same point, but their centres of mass coincide. If they do not coincide, it will give rise to an electric dipole or a polar molecule.
“WHAT IS THE MAGNITUDE OF ELECTRIC FIELD IN THE GAUSSIAN SURFACE OF A CUBE,AT ITS FACE ,AT ITS CENTRE, AT ITS CORNERS OR AT ANY OTHER POINT INSIDE THE CUBE. ALSO TELL ME THE THE WAY TO KNOW IT.”
According to Gauss Theorem, the total electric flux through a closed surface,
The Gaussian surface is an imaginary surface. So, for calculating the electric field at a point using Gauss theorem, we have to imagine a Gaussian surface symmetric with the given charge distribution.
I have assumed that there is a point charge Q at the centre of the cube.
At its face (at a point on the face on the line radially outwards from the point charge at the centre of the cube and perpendicular to the face)
The distance is equal to half the length of side of the cube (Say L).
On the corner,
Calculate the distance from the centre of the cube to its corner and replace (L/2) in the above equation with that distance.