Answer:
The speed of ejection is 
Solution:
As per the question:
Magnetic field density, B = 0.4 T
Density of the material in the sunspot, 
Now,
To calculate the speed of ejection of the material, v:
The magnetic field energy density is given by:
This energy density equals the kinetic energy supplied by the field.
Thus
where
m = mass of the sunspot in 
= 
Answer:
E=0
Explanation:
Electric field due to each thin sheet of charge=\sigma/2\varepsilon
let us say the right plate has positive charge density \varepsilonand left sheet has a negative charge density -\varepsilon .
In the region between the plates,the electric field due to each plate is in same direction,
E=\sigma/2\varepsilon-(-\sigma/2\varepsilon)
E=\sigma/\varepsilon
in the region outside the plates, the field due to the plates is in opposite directions
E=-\sigma/2\varepsilon-(-\sigma/2\varepsilon)
E=-\sigma/2\varepsilon+\sigma/2\varepsilon
E=0
First, torque is equal to force times the distance. for the first force that is applied, the torque is zero because is applied at the hinge. so the net torque:
t = ( 12 N ) ( 0 m ) ( cos 30 ) + ( 12 N ) ( 1.68 m ) cos 45
t = 14.26 Nm is the torque with respect to the hinge
D = V0t + 0.5at^2
Where d is the distance
V0 is the initial velocity
A is the acceleration
T is time
From the graph a = 4/3 m/s2
D = 0(3) + 0.5( 4/3 m/s2) ( 3 s)^2
D = 6 m
