A stream of water flowing horizontally

Volume of water striking the wall per second = v x A = 15 x 10 -2 = 0.15 m -3 s -1 Mass of water hitting wall per second = 9 x v x A = 1000 x 0.15 kg s -1 = 150 kg s -1 Initial momentum of water per second = 150 x 150 = 2250 kg ms -1 Final momentum of water per second = 0 (There is no rebound of water) Magnitude of force = Rate of change of momentum = 2250 N

A helicopter of mass 1000 k g rises with a vertical acceleration of 15 m / s 2. The crew and the passengers weigh 300 k g. Give the magnitude and direction of the(a) Force on the floor by the crew and passengers.(b) Action of the rotor of the helicopter on the surrounding air.(c) Force on the helicopter due to the surrounding air. Views: 5,807 70 k g stands on a weighing scale in a lift which is moving. (a) Upwards with a uniform speed of 10 m s − 1. (b) Downwards with a uniform acceleration of 5 m s − 2. (c) Upwards with a uniform acceleration of 5 m s − 2. What would be the reading on the scale in each case? (d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity? Views: 5,536 θ = 1 8 ∘ 5 6 ′ with the vertial towards the West. Example 9 A person standing on a road has to hold his umbrella at 6 0 ∘ with the vertical to keep the rain away. He throws the umbrella and starts running at 20 ms − 1. He finds that rain drops are hitting his head vertically. Find the speed of the rain drops with respect to (a) the road (b) the moving person. Solution. When person is at rest with respect to ground, the rain is coming to him at an angle 6 0 ∘ witt the vertical i.e. along OB, with velocity 20 ms − 1 A stream of water flowing horizontally with a speed of 15 m s − 1 gushes out of a tube of cross-sectional area 1 0 − 2 m 2, and hits a vertical wall nearby. What is the force exerted on the wall by the impact of water, assug it does not rebound...

This problem has been solved! You'll get a detailed solution from a subject matter expert that helps you learn core concepts. See Answer See Answer See Answer done loading Question:-1 2 A stream of water flowing horizontally with a speed of 15 m s pushes out of a tube of cross-sectional area 10-2 m², and hits a vertical wall nearby. What is the force exerted on the wall by the impact of water, assuming it does not rebound ?

Here , ` upsilon = 15 ms^(-1) ` Area of cross section , ` a = 10^(-2) m^(2) , F = ? ` Volume of water pushing out/sec `= a xx upsilon = 10^(-2) xx 15 m^(3) s^(-1)` As density of water is `10^(3) kg //m^(-3)` , thereforce mass of water striking the wall per sec `m = (15 xx 10^(-2)) xx 10^(3) = 150 kg//s` As `F = ("change in linear momentum")/("time") :. F = (m xx upsilon)/ (t) = (150 xx 15)/(1) = 2250 N`.

Flowing water speed, v=15 m/s A= 10-2 m2 is the cross-sectional area of the tube. The distance travelled in one second is equal to the velocity v. 1000 kgm-3 is the density of water. ρ x A x v = 1000 kgm -3 x 10 -2 m 2 x 15 m/s = 150 kg/s The force exerted on the wall due to the impact of water equals the water’s momentum loss per second. = Weight x Speed 2250 N = 150 kg/s x 15 m/s

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Torricelli's law describes the parting speed of a jet of water, based on the distance below the surface at which the jet starts, assuming no air resistance, viscosity, or other hindrance to the fluid flow. This diagram shows several such jets, vertically aligned, leaving the reservoir horizontally. In this case, the jets have an Torricelli's law, also known as Torricelli's theorem, is a theorem in v Torricelli's law can only be applied when viscous effects can be neglected which is the case for water flowing out through orifices in vessels. Experimental verification: Spouting can experiment [ ] Every physical theory must be verified by experiments. The spouting can experiment consists of a cylindrical vessel filled up with water and with several holes in different heights. It is designed to show that in a The outflowing jet forms a downward parabola where every parabola reaches farther out the larger the distance between the orifice and the surface is. The shape of the parabola y ( x ) two quantities could be responsible for this discrepancy: the outflow velocity or the effective outflow cross section. In 1738 Daniel Bernoulli attributed the discrepancy between the theoretical and the observed outflow behavior to the formation of a A A Clepsydra problem [ ] A Alternatively, by carefully selecting the shape of the vessel, the water level in the vessel can be made to decrease at constant rate. By measuring the level of water remaining in the vessel, the time can be measur...