Answer:
towards the south
Explanation:
When the electron enters the region with magnetic field, it experiences a magnetic force perpendicular to both the directions of the electron's velocity and the magnetic field.
The direction of the force exerted on the electron can be found by using the right-hand rule:
- Index finger: direction of the velocity of the electron --> towards the east
- Middle finger: direction of the magnetic field --> downward
- Thumb: direction of the force on a positive particle --> towards the north
However, the electron is a negatively charged particle, so we must reverse the direction of the force: therefore, the force exerted on the electron is towards the south.
As we know that electrostatic force is a conservative force
so we can say by the condition of conservative force

here we can rearrange the above equation as

now integrate both sides

Now we know by the definition of work done by a force is given by

now work done by conservative force is given as

Now from above work done by electric field to move charge from one point to other is given as

so here work done is given as

so change in potential energy is given by work done
Explanation:
Total surface area of cylinder:

sheet of metal required = 120.95 cm^2
Answer:
<em>The rebound speed of the mass 2m is v/2</em>
Explanation:
I will designate the two masses as body A and body B.
mass of body A = m
mass of body B = 2m
velocity of body A = v
velocity of body B = -v since they both move in opposite direction
final speed of mass A = 2v
final speed of body B = ?
The equation of conservation of momentum for this system is
mv - 2mv = -2mv + x
where x is the final momentum of the mass B
x = mv - 2mv + 2mv
x = mv
to get the speed, we divide the momentum by the mass of mass B
x/2m = v = mv/2m
speed of mass B = <em>v/2</em>
Explanation:
a) We need to write down first Newton's 2nd law as applied to the given system. The equations of motion for the x- and y-axes can be written as follows:


From Eqn(2), we see that

so using Eqn(3) on Eqn(1), we get

Solving for the acceleration, we see that


b) Now that we have the acceleration, we can now solve for the velocity of the crate at the bottom of the plane. Using the equation

Since the crate started from rest,
Thus our equation reduces to


