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Given, 

Mass of the helicopter, m_h = 1000 kg 

Mass of the crew and passengers, m_p = 300 kg

Total mass of the system, m = 1300 kg

Acceleration of the helicopter, a = 15 m/s² 

Using Newton’s second law of motion, the reaction force R,

           R – m_pg = ma

                          = m_p(g + a)

                          = 300 (10 + 15) 

                          = 300 × 25 

                          = 7500 N 

The reaction force will be directed upwards, the helicopter is accelerating vertically upwards. 

According to Newton’s third law of motion, the force on the floor by the crew and passengers  = 7500 N, directed downward.

(b)

Using Newton’s second law of motion, the reaction force R’ experienced by the helicopter can be calculated as, 

                 R' - mg = ma 

                             = m(g + a)

                             = 1300 (10 + 15)

                             = 1300 × 25 
                                     
                             = 32500 N 

The reaction force experienced by the helicopter from the surrounding air is acting upward. Hence, as per Newton’s third law of motion, the action of the rotor on the surrounding air will be 32500 N, directed downward.

(c) The force on the helicopter due to the surrounding air is 32500 N, directed upwards. 

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The free body diagram of the stone at the lowest point is shown in the figure below:


                   


According to Newton’s second law of motion, the net force acting on the stone at this point is equal to the centripetal force.

i.e.,  F_net = T - mg =                     ...(i)
where,

v₁ is the velocity at the lowest point. 

The free body diagram of the stone at the highest point is shown in the following figure. 

                 

Using Newton’s second law of motion, 


        T + mg =                        ...(ii)

where, v₂ is the velocity at the highest point.
Net force acting at the lowest = (T - mg)

Net force at the highest points = (T + mg) 

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Given,
                  

∴                     

We have,

Velocity of the aircraft, v = 720 km/hr = 200 m/s

 

∴ Radius of the loop, 

                                 = 15232.86 m

                                 = 15.23 km

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Weight of one coin = mg

Weight of three coins = 3mg

Hence, the force exerted on the 7^th coin by the three coins on its top is 3mg.

This force acts vertically downward.

(b) Force on the seventh coin by the eighth coin is because of the weight of the eighth coin and the other two coins (ninth and tenth) on its top.

Weight of the eighth coin = mg

Weight of the ninth coin = mg

Weight of the tenth coin = mg

Total weight of these three coins = 3mg 

Hence, the force exerted on the 7^th coin by the eighth coin is 3mg.

This force acts vertically downward.

(c) The 6^th coin experiences a downward force because of the weight of the four coins (7^th, 8^th, 9^th, and 10^th) on its top. 

Therefore, the total downward force experienced by the 6^thcoin is 4mg. 

As per Newton’s third law of motion, the 6^th coin will produce an equal reaction force on the 7^th coin, but in the opposite direction.

Hence, the reaction force of the 6^th coin on the 7^thcoin is of magnitude 4mg.

This force acts in the upward direction. 

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Cross-sectional area of the tube, A = 10^–2 m²

Volume of water coming out from the pipe per second,

                 V = Av 

                     = 15 × 10^–2 m³/s 

Density of water, ρ = 10³ kg/m³ 

Mass of water flowing out through the pipe per second = ρ ×V 

           = 150 kg/s

The water strikes the wall and does not rebound.

Therefore, according to Newton's second law of motion, 

Force exerted by the water on the wall,

F = Rate of change of momentum = 

                                                    = 

                                                    = 150 × 15

                                                    = 2250 N