Answer:
Part a)
V = 18.16 V
Part b)

Part c)
P = 672 Watt
Part d)
V = 5.84 V
Part e)

Explanation:
Part a)
When battery is in charging mode
then the potential difference at the terminal of the cell is more than its EMF and it is given as

here we have



now we have

Part b)
Rate of energy dissipation inside the battery is the energy across internal resistance
so it is given as



Part c)
Rate of energy conversion into EMF is given as



Now battery is giving current to other circuit so now it is discharging
now we have
Part d)



Part e)
now the rate of energy dissipation is given as



Answer:
The appropriate solution is "61.37 s".
Explanation:
The given values are:
Boat moves,
= 10 m/s
Water flowing,
= 1.50 m/s
Displacement,
d = 300 m
Now,
The boat is travelling,
= 
= 
Travelling such distance for 300 m will be:
⇒ 

On putting the values, we get


Throughout the opposite direction, when the boat seems to be travelling then,
= 
= 
Travelling such distance for 300 m will be:
⇒ 

On putting the values, we get


hence,
The time taken by the boat will be:
= 
= 
A. Condensation
Hope this helps!!!
Answer:
(a) Ferromagnet
Explanation:
Ferromagnetism is defined as the property by which certain magnets form the permanent magnets.
It is tone of the strong magnetism and it is common phenomenon of magnet in the everyday life of magnetism.
Permanent magnets are made up of ferromagnetic material, in this if the magnetic field is applied then this material is magnetized but do not losses its magnetic property after removal of external magnetic field.
Answer:
The angle of separation is
Explanation:
From the question we are told that
The angle of incidence is 
The refractive index of violet light in diamond is 
The refractive index of red light in diamond is 
The wavelength of violet light is
The wavelength of red light is
Snell's Law can be represented mathematically as

Where
is the angle of refraction
=> 
Now considering violet light

substituting values




Now considering red light

substituting values




The angle of separation between the red light and the violet light is mathematically evaluated as

substituting values

