Answer:
The time interval is 
Explanation:
From the question we are told that
The constant acceleration is 
The displacement is
According to the second equation of motion we have that
given that the blade started from rest
which is the initial angular velocity is 0
So
=> 
substituting values
=> 
=> 
Table sugar dissolves in water because when a sucrose molecule breaks from the sugar crystal, it is immediately surrounded by water molecules. The sucrose has hydroxyl groups that have a slight negative charge. ... Sand can't dissolve in waterbecause the 'spaces' in between the water particles. :)
Answer:

Explanation:
Magnetic field due to straight long wire is always perpendicular to the line joining the wire and the point
it is given by the equation

now the magnetic field due to 30 A current is given as


Now magnetic field due to other wire having current 40 A is given as


now net field along Y direction is given as


now for X direction magnetic field we know



Now net magnetic field is given as


