KNOWLEDGE + POINT(gyanni kitab)

 THOMSON'S MODEL OF AN ATOM Hello, everyone . Welcome in my new blog.  In this blog I have explained the model of atom given by Thomson.                       Thomson proposed the model of an atom to be similar to that of a Christmas pudding. The electrons, in a sphere of positive charge, were like currants (dry fruits) in a spherical Christmas pudding. Thomson proposed that: (i) An atom consists of a positively  charged sphere and the electrons are  embedded in it. (ii) The negative and positive charges are  equal in magnitude. So, the atom as a  whole is electrically neutral.  Although Thomson’s model explained that  atoms are electrically neutral, the results of  experiments carried out by other scientists  could not be explained by this model, as we  will see below. , THANKS FOR COMING

  What is   NEWTON'S THIRD LAW OF MOTION  ? It states that every action (force applied) has an equal and opposite reaction. EXAMPLE FOR UNDERSTANDING :- Suppose, here the force exerted by the hands on the walls be 10 N . So, according to 3rd law of motion action is equals to a reactions and here the action is of 10 N , so the reactions will also be 10 N. ACTION = REACTION 10 N = 10 N APPLICATION OF 2nd LAW OF MOTION 1. WALKING Newton's third law of motion is also applicable for a simple activity of walking. During walking we push the ground backwards or apply a force in backward direction which is the action and then we get an reaction force and se we are able to move forward. THANKS WAS VIEWING COMMING SOON WITH A  NEW EXPERITENT BLOG FOR KNOLEDGE OF NEWTON'S 2nd LAW OF MOTION CLIK HERE ----  Experiment 1

  What  is NEWTON'S SECOND LAW OF MOTION ?  It states that Rate of change in momentum is directly proportional to the applied force on an object , or we can say Force is the product of mass and acceleration.                   F = ma MATHEMATICAL DERIVATION F  ∝ rate of change of momentum F  ∝ change in momentum / time  let the mass of the object is to be m and initial velocity and final velocity is u , v respectively so the initial momentum and final momentum will be mu and mv respectively and so the change in momentum is mv - mv in time t . Now, F  ∝ mv - mu / t F  ∝ m(v-u) / t since,  we know v-u / t is equals to acceleration 'a'. F  ∝ ma now, removing the proportionality symbol we have  F = kma (k is the constant for  ∝ ) now if the mass and acceleration are 1 then value of k is equals to 1 Therefore , F = ma            ...