Explanation :
Dispersion forces are also known as London dispersion forces. It is the weakest force. Also, it is the part of the Van der Waals forces.
(1) This force is exhibited by all atoms and molecules.
(2) These forces are the result of the fluctuations in the electron distribution within molecules or atoms. Due to these fluctuations, the electric field is created. The magnitude of this force is explained in terms of Hamaker constant 'A'.
(3) Dispersion forces result from the formation of instantaneous dipoles in a molecule or atom. When electrons are more concentrated in a place, instantaneous dipoles formed.
(4) Dispersion force magnitude depends on the amount of surface area available for interactions. If the area increases, the size of the atom also increase. As a result, stronger dispersion forces.
So, the false statement is "Dispersion forces always have a greater magnitude in molecules with a greater molar mass".
In the offensive role, the players try to get a goal.
In the defensive roll, The players try to protect the goal
Hoped this helped a little :)
Answer:
(a) 1.58 V
(b) 0.0126 Wb
(c) 0.0493 V
Solution:
As per the question:
No. of turns in the coil, N = 400 turns
Self Inductance of the coil, L = 7.50 mH =
Current in the coil, i =
A
where

Now,
(a) To calculate the maximum emf:
We know that maximum emf induced in the coil is given by:

![e = L\frac{d}{dt}(1680)cos[\frac{\pi t}{0.0250}]](https://tex.z-dn.net/?f=e%20%3D%20L%5Cfrac%7Bd%7D%7Bdt%7D%281680%29cos%5B%5Cfrac%7B%5Cpi%20t%7D%7B0.0250%7D%5D)
![e = - 7.50\times 10^{- 3}\times \frac{\pi}{0.0250}\times \frac{d}{dt}(1680)sin[\frac{\pi t}{0.0250}]](https://tex.z-dn.net/?f=e%20%3D%20-%207.50%5Ctimes%2010%5E%7B-%203%7D%5Ctimes%20%5Cfrac%7B%5Cpi%7D%7B0.0250%7D%5Ctimes%20%5Cfrac%7Bd%7D%7Bdt%7D%281680%29sin%5B%5Cfrac%7B%5Cpi%20t%7D%7B0.0250%7D%5D)
For maximum emf,
should be maximum, i.e., 1
Now, the magnitude of the maximum emf is given by:

(b) To calculate the maximum average flux,we know that:

(c) To calculate the magnitude of the induced emf at t = 0.0180 s:


Answer:
The mass of the rule is 56.41 g
Explanation:
Given;
mass of the object suspended at zero mark, m₁ = 200 g
pivot of the uniform meter rule = 22 cm
Total length of meter rule = 100 cm
0 22cm 100cm
-------------------------Δ------------------------------------
↓ ↓
200g m₂
Apply principle of moment
(200 g)(22 cm - 0) = m₂(100 cm - 22 cm)
(200 g)(22 cm) = m₂(78 cm)
m₂ = (200 g)(22 cm) / (78 cm)
m₂ = 56.41 g
Therefore, the mass of the rule is 56.41 g