Dispersion im pretty sure
Answer:
The charge on the dust particle is 
Explanation:
From the question we are told that
The length is 
The width is 
The charge is 
The mass suspended in mid-air is 
Generally the electric field on the carpet is mathematically represented as

Where
is the permittivity of free space with value 
substituting values


Generally the electric force keeping the dust particle on the air equal to the force of gravity acting on the particles

=> 
=> 
=> 
=> 
Answer:

Explanation:
From the question we are told that:
Weight 
Mass of Weight 

Generally the acceleration due to gravity on the Moon is one-sixth that on Earth.
Therefore
The equation for Weight on Moon is given as



Answer: The ice absorb 6671.1 kJ of thermal energy.
Explanation:
The conversions involved in this process are :


Now we have to calculate the enthalpy change.

where,
= enthalpy change = ?
m = mass of ice = 20.0 kg =
(1kg=1000g)
n = number of moles of ice= 
= enthalpy change for fusion = 6.01 KJ/mole = 6010 J/mole
Now put all the given values in the above expression, we get

(1 kJ = 1000 J)
Therefore, the enthalpy change is, 6671.1 kJ