a) For the motion of car with uniform velocity we have , 
, where s is the displacement, u is the initial velocity, t is the time taken a is the acceleration.
In this case s = 520 m, t = 223 seconds, a =0 
Substituting
The constant velocity of car a = 2.33 m/s
b) We have 
s = 520 m, t = 223 seconds, u =0 m/s
Substituting
Now we have v = u+at, where v is the final velocity
Substituting
v = 0+0.0209*223 = 4.66 m/s
So final velocity of car b = 4.66 m/s
c) Acceleration = 0.0209
The air that is inside a ship is much less dense than water. That's what keeps it floating! ... The closer the total density of the ship is to the density of the same volume of water, the greater the amount of the ship that will be in the water.
The correct answer to the question is vertically downward i.e towards the centre of earth.
EXPLANATION:
As per the question, the box is pulled to the right.
Hence, the direction of the applied force is towards right.
We are asked to determine the direction of the gravitational force that acts on the body.
Before answering this question, first we gave to understand the gravitational force of earth.
Any body present on the surface of earth is attracted with the force of gravity of earth ( gravitational force ) towards its centre. It is equivalent to the weight of the body.
The force of gravity is always directed towards the centre of earth irrespective of the nature of applied force.
Hence, the direction of the gravitational force which acts on the box is vertically downward.
Answer:
The atmospheric pressure and boiling point are directly proportional
Increasing atmospheric pressure increases the boiling point also
Explanation:
The atmosphere contain molecules that are in constant motion. They exert a downward force on a liquid’s surface. The higher the air pressure, the harder it is for the liquid to evaporate. Therefore, the boiling point of a solvent or liquid is affected by the atmospheric pressure and boiling point is raised.
A liquid in a high pressure environment boils at a higher temperature.
When placed in a lower pressure environment it boils at a lower temperature.
The number of charge drifts are 3.35 X 10⁻⁷C
<u>Explanation:</u>
Given:
Potential difference, V = 3 nV = 3 X 10⁻⁹m
Length of wire, L = 2 cm = 0.02 m
Radius of the wire, r = 2 mm = 2 X 10⁻³m
Cross section, 3 ms
charge drifts, q = ?
We know,
the charge drifts through the copper wire is given by
q = iΔt
where Δt = 3 X 10⁻³s
and i = 
where R is the resistance
R = 
ρ is the resistivity of the copper wire = 1.69 X 10⁻⁸Ωm
So, i = 
q = 
Substituting the values,
q = 3.14 X (0.02)² X 3 X 10⁻⁹ X 3 X 10⁻³ / 1.69 X 10⁻⁸ X 0.02
q = 3.35 X 10⁻⁷C
Therefore, the number of charge drifts are 3.35 X 10⁻⁷C
