A container of water has a cross-sectional area of \(A = 0.3 \, \text{m}^2\). A piston sits on top of the water. There is a spout located 0.15 m from the bottom of the tank, open to the atmosphere, and a stream of water exits the spout. The cross-sectional area of the spout is \(A_s = 4.6 \times 10^{-4} \, \text{m}^2\).An illustration shows a vertical container filled with water. A 20 kg mass sits on a piston at the top, applying pressure to the water. The piston's area is the same as the container's cross-section. The spout, located 0.15 m from the bottom, allows water to exit. The container's height from the piston to the bottom is 0.5 m. The density of water is given as \(\rho = 1000 \, \text{kg/m}^3\).Questions:(a) What is the speed of the water (in m/s) as it leaves the spout?(b) If the opening of the spout is located 2.5 m above the ground, how far from the spout does the water hit the floor? Ignore all friction and dissipative forces. (Enter the horizontal distance from the spout in meters.)The image consists of a transparent cylinder with water, a piston on top, and a spout on the side. The spout is positioned above ground level, allowing analysis of water flow using principles like Bernoulli’s equation and projectile motion.

Name: A container of water has a cross-sectional area of \(A = 0.3 \, \text
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Description: A container of water has a cross-sectional area of \(A = 0.3 \, \text{m}^2\). A piston sits on top of the water. There is a spout located 0.15 m from the bottom of the tank, open to the atmosphere, and a stream of water exits the spout. The cross-se

(a) Write Bernoulli's equation at height h1 (piston water interface). P1+12ρv2+ρgh1=cont. 1 Write Bernoulli's equation at the height of h2. P2+12ρv2+ρgh2=cont. 2 Compare both these expressions for the velocity of water at the spout. P1+12ρv2+ρgh1=P2+12ρv2+ρgh2P1+0+ρgh1=P2+12ρv2+ρgh22P1-P2+ρgh1-h2ρ=v2v=2P1-P2+ρgh1-h2ρ 3 The pressure difference is given by P1-P2=Patm+F/A-Patm=F/A Substitute the value of pressure difference in equation (3). v=2F/A+ρgh1-h2ρ=2mg/Aρ+gh1-h2=220 kg9.80 m/s2/0.3 m21000 kg/m3+9.80 m/s20.5 m-0.15 m=2.86 m/s (b) Let's calculate how long the water stream takes to reach the ground using a kinematic equation. h=v0t+12gt2=0+12gt2t=2hg=22.5 m9.80 m/s2=0.714 sAs the speed of the water stream is the same at the starting and endpoints of the spout. The distance of the water stream from the tank is, d=vt=2.86 m/s0.714 s=2.04 m

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A container of water has a cross-sectional area of A = 0.3 m2, A piston sits on top of the water (see the following figure). There is a spout located 0.15 m from the bottom of the tank, open to the atmosphere, and a stream of water exits the spout. The cross sectional area of the spout is A,- 4.6 x 10-4 m2. 20 kg p= 1000 O m3 0.5 m 0.15 m (a) What is the speed of the water (in m/s) as it leaves the spout? m/s (b) If the opening of the spout is located 2.5 m above the ground, how far from the spout does the water hit the floor? Ignore all friction and dissipative forces. (Enter the horizontal distance from the spout in meters.) m

A container of water has a cross-sectional area of A = 0.3 m2, A piston sits on top of the water (see the following figure). There is a spout located 0.15 m from the bottom of the tank, open to the atmosphere, and a stream of water exits the spout. The cross sectional area of the spout is A,- 4.6 x 10-4 m2. 20 kg p= 1000 O m3 0.5 m 0.15 m (a) What is the speed of the water (in m/s) as it leaves the spout? m/s (b) If the opening of the spout is located 2.5 m above the ground, how far from the spout does the water hit the floor? Ignore all friction and dissipative forces. (Enter the horizontal distance from the spout in meters.) m

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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A container of water has a cross-sectional area of \(A = 0.3 \, \text{m}^2\). A piston sits on top of the water. There is a spout located 0.15 m from the bottom of the tank, open to the atmosphere, and a stream of water exits the spout. The cross-sectional area of the spout is \(A_s = 4.6 \times 10^{-4} \, \text{m}^2\).

An illustration shows a vertical container filled with water. A 20 kg mass sits on a piston at the top, applying pressure to the water. The piston's area is the same as the container's cross-section. The spout, located 0.15 m from the bottom, allows water to exit. The container's height from the piston to the bottom is 0.5 m. The density of water is given as \(\rho = 1000 \, \text{kg/m}^3\).

Questions:
(a) What is the speed of the water (in m/s) as it leaves the spout?
(b) If the opening of the spout is located 2.5 m above the ground, how far from the spout does the water hit the floor? Ignore all friction and dissipative forces. (Enter the horizontal distance from the spout in meters.)

The image consists of a transparent cylinder with water, a piston on top, and a spout on the side. The spout is positioned above ground level, allowing analysis of water flow using principles like Bernoulli’s equation and projectile motion.

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