Formula: The moment of a force M = F × d
Given: F = 40 N and d = 90 cm = 0.9 m.
Hence, moment of the force = 40 × 0.9 = 36 N m.

i) Dimensions of the heavenly bodies can be measured using the gravitation . Mass of the Earth, radius of the Earth, acceleration due to gravity, etc. can be calculated with a higher accuracy.
ii) It helps in discovering new stars and planets.
iii) One of the irregularities in the motion of stars is called 'Wobble' lead to the disturbance in the motion of a planet nearby. In this condition the mass of the star can be calculated using the law of gravitation.
iv) Helps to explain germination of roots is due to the property of geotropism which is the property of a root responding to the gravity.
v) Helps to predict the path of the astronomical bodies.

Application of Newton's law of gravitation
(i) Dimensions of the heavenly objects can be measured using gravitation law. Mass of the earth, radius of the earth, acceleration due to gravity etc. can be calculated with a higher accuracy.
(ii) Helps in discovering new stars and planets. Mass of the double stars can be calculated.
(iii) One of the irregularities in the motion of stars is called "Wobble" which leads to the disturbance in the motion of planet nearby. In this condition mass of the star can be calculated using law of gravitation.
(iv) Helps to explain germination of roots due to the property of geotropism, which is the property of root responding to the gravity.
(v) Helps to predict the path of the astronomical bodies.

(i) Propulsion of rockets is based on law of conservation of linear momentum as well as Newton's III law of motion.
(ii) Rockets are filled with a fuel (either liquid or solid) in the propellant tank.
(iii) When the rocket is fired, this fuel is burnt and a hot gas is ejected with high speed from the back nozzle producing a huge momentum.
(iv) To balance this momentum, an equal and opposite reaction force is produced combustion chamber which makes the rocket project forward.
(v) While in motion, the mass of the rocket gradually decreases, until the fuel is completely burnt out.
(vi) Since there is no net external force acting on it, the linear momentum of the system is conserved.
(vii) The mass of the rocket decreases with altitude, which results in the gradual increase in velocity of the rocket.
(viii) At one stage, it reaches a velocity, which is sufficient to just escape from the gravitational pull of the Earth. This velocity is called escape velocity.

Given: K₁ = K₂ = K, m₁ = 4m₂
The kinetic energy of the truck \(=\frac{1}{2} m_{1} v_{1}^{2} \Rightarrow v_1= \sqrt\frac{2 k}{m_{1}}\)
The kinetic energy of the bike \(=\frac{1}{2} \mathrm{~m}_{2} \mathrm{v}_{2}^{2}\Rightarrow v_2= \sqrt\frac{2 k}{m_{2}}\)
∴ Momentum p = mv
∴ Momentum of the two bodies are given by,
\(P_1=\sqrt{2m_1K,}\)
\(P_2=\sqrt{2m_2K,}\)
\(\therefore \frac{P_1}{P_2} =\sqrt{\frac{2 m_1K}{2m_{2}K}}=\sqrt{\frac{m_1}{m_{2}}}=\sqrt{\frac{4m_2}{m_{2}}}=\frac{\sqrt { 4}}{\sqrt 1}= \frac{2}{1}\)
Ratio of momenta = 2: 1