Explanation:
Rutherford proposed a revised model for the atom, called the planetary model. The previous model of the atom was Thomson's Plum Pudding Model which consisted of freely moving positive and negative charges inside the atom.
Rutherford proposed his model after an experiment he conducted called the Gold Foil Experiment. This experiment consisted of a thin gold sheet into which alpha particles were shot upon and they were detected by a sensor. The image attached will give a better explanation of this. In this experiment he shot a beam of alpha particles(helium nucleus) at a thin sheet of gold. Rutherford hypothesised that there should be minimum deflection of the positively charged alpha particles occuring due to the repulsion of the alpha particle with the positive charges in the thin gold sheet. This was not the case.
However what he found was that most of alpha particles went straight through the thin sheet of gold but some were reflected back to him. This surprised him. Hence he proposed that most of the atom must be empty space as most of the alpha particles went straight through the sheet and there must be a heavy nucleus inside the atom causing the alpha particles to bounce back.
Answer:
Explanation:
Electron's kinetic energy = 2 eV
= 2 x 1.6 x 10⁻¹⁹ J
1/2 m v² = 3.2 x 10⁻¹⁹
1/2 x 9.1 x 10⁻³¹ x v² = 3.2 x 10⁻¹⁹
v² = .703 x 10¹²
v = .8385 x 10⁶ m/s
Electrons revolve in a circular orbit when forced to travel in a magnetic field whose radius can be expressed as follows
r = mv / Bq
where m , v and q are mass , velocity and charge of electron .
here given magnetic field B = 90 mT
= 90 x 10⁻³ T
Putting these values in the expression above
r = mv / Bq
= 
= .052 mm.
The answer is A. the fields lines never cross, if you bring another magnet near it, the lines work just compress
Explanation:
Given that,
Mass if the rock, m = 1 kg
It is suspended from the tip of a horizontal meter stick at the 0-cm mark so that the meter stick barely balances like a seesaw when its fulcrum is at the 12.5-cm mark.
We need to find the mass of the meter stick. The force acting by the stone is
F = 1 × 9.8 = 9.8 N
Let W be the weight of the meter stick. If the net torque is zero on the stick then the stick does not move and it remains in equilibrium condition. So, taking torque about the pivot.

W = 3.266 N
The mass of the meters stick is :

So, the mass of the meter stick is 0.333 kg.