Answer:
Explanation:
1. False
The force you apply on crate is equal and opposite to the force that crate applies on you by Newton's third law of motion.
The force must over come the static frictional force between the crate and the floor.
2. True
The object can move along another direction than the direction of net force. For example, when a car slows down, the net force is opposite to the direction of motion.
3. True
An object moving at constant velocity has zero net force acting on it.
4. False
An object at rest has forces acting on it but the summation of all the forces is zero i.e. the net force is zero.
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:
n = 5 approx
Explanation:
If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back
= e ( coefficient of restitution ) = 
and

h₁ is height up-to which the ball bounces back after first bounce.
From the two equations we can write that


So on

= .00396
Taking log on both sides
- n / 2 = log .00396
n / 2 = 2.4
n = 5 approx
Answer:
Explanation:
Force on a moving charge is given by the following relation
F = q ( v x B )
for proton
q = e , v = vi , B = Bk
F = e ( vi x Bk )
= Bev - j
= - Bevj
The direction of force is along negative of y axis or -y - axis.
for electron
q = - e , v = vi , B = Bk
F = - e ( vi x Bk )
= - Bev - j
= Bevj
The direction of force is along positive of y axis or + y - axis.