A camera drone of mass 1.20 kg hovers at a fixed point above the ground. The drone has fourpropellers.propellercameraIn a time of 1.00 s, each propeller pushes a mass of 0.400 kg of air vertically downwards.Assume that the air above the propellers is stationary.What is the speed of the air leaving each propeller?A 0.750 m s –1 B 3.00 m s –1 C 7.36 m s –1 D 29.4 m s –1

To solve this problem, we need to use the principle of conservation of momentum. The momentum of the drone is zero (since it's hovering, not moving), so the total momentum of the air must also be zero.

The momentum of each mass of air as it leaves the propeller is its mass times its velocity (p=mv). Since there are four propellers, the total momentum of the air is 4 times the momentum of the air from one propeller.

Let's denote the speed of the air leaving each propeller as v. So, the total momentum of the air is 40.400 kgv = 1.6 kg*v.

This must equal the total momentum of the drone, which is its mass times its velocity. Since the drone is hovering, its velocity is zero, so its momentum is 1.20 kg * 0 = 0.

Setting these two equal gives us 1.6 kg*v = 0. Solving for v gives us v = 0 / 1.6 kg = 0 m/s.

However, this is not one of the answer choices, which suggests that there may be a mistake in the problem or in our understanding of it. The drone must exert a downward force on the air to counteract the force of gravity and stay aloft, so the air should be moving downwards, not stationary.

If we assume that the drone exerts a force equal to its weight (mass times gravity) on the air, then the momentum of the air should be equal to the drone's weight times the time over which the force is applied. This gives us a momentum of 1.20 kg * 9.81 m/s^2 * 1.00 s = 11.77 kg*m/s.

Setting this equal to the total momentum of the air gives us 1.6 kgv = 11.77 kgm/s. Solving for v gives us v = 11.77 kg*m/s / 1.6 kg = 7.36 m/s.

So, the speed of the air leaving each propeller should be 7.36 m/s, which corresponds to answer choice C.