Part 3 of 4 - AnalyzeIn the figure, take location O at the top surface and location 2 at the exiting stream. Assuming the top isopen to the atmosphere, then P, = Po. At location 2, by Newton's third law, the water must push on the airjust as strongly as the air pushes on the water, so P2 = Po-(a) By Bernoulli's law, we haveP」 + 글pri2+ pgy1= P2 + 극p22 + pgy2Here, A, » A2, so v, « v2. Assuming v, = 0 and P, = P, = Po, gives the following.V2 = v 2g(y1 – 2)=V 2 (9.80 m/s)(17.9 m/s16.316.3 m= 17.874Part 4 of 4 - AnalyzeWe identify the measured flow rate asA2V2= [2.7x 10-3 m3/min.The formula for flow rate is as follows.nd?V21and solving for the diameter, we have12.7v x 10-3 m/min)4d =IT V22.7x 10-3 m/min) 20.0566 Xm/s) (60 s/1 min=17.9Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. x 10-3mmm. A large storage tank, open at the top and filled with water, develops a small hole in its side at a point 16.3 mbelow the water level. The rate of flow from the leak is found to be 2.70 x 103 m3/min.(a) Determine the speed at which the water leaves the hole.(b) Determine the diameter of the hole.1Part 1 of 4 - ConceptualizeAn object falling freely from height h from rest reaches a speed v = V 2gh. For ideal flow, we can expect thissame speed for the spurting water. A few millimeters seem reasonable for the diameter of a hole on the sideof a tank that sprays out only a couple of liters every minute.Part 2 of 4 - CategorizeWe use Bernoulli's equation and the expression for volume flow rate contained in the equation of continuity forfluids.Part 3 of 4 - AnalyzeIn the figure, take location O at the top surface and location 2 at the exiting stream. Assuming the top isopen to the atmosphere, then P, = Po. At location 2, by Newton's third law, the water must push on the airjust as strongly as the air pushes on the water, so P, = Po:(a) By Bernoulli's law, we haveP, +pv,? + pgy, = P2 + pv² + pgy2:

Please see the answer in step 2

Part 3 of 4 - Analyze In the figure, take location O at the top surface and location at the exiting stream. Assuming the top is open to the atmosphere, then P, = Po- At location 0, by Newton's third law, the water must push on the air just as strongly as the air pushes on the water, so P, = Po- (a) By Bernoulli's law, we have Here,A, » Ag, s0 V, « V3. Assuming v, = 0 and P, = P2 = Por gives the following. V2 =v 2g(y, - Y2) 2 (9.80 m/s?)(16.3 16.3 m - 17.874 17.9 m/s Part 4 of 4- Analyze We identify the measured flow rate as AgVz=27 x 10 m/min. The formula for flow rate is as follows. (na² and solving for the diameter, we have 2.7 | × 10-³ m³/min) ( 2.7 v × 10-³ m³/min) -0.0566 X 17.9 m/s) (60 s/1 min Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. x 103 m = mm.

Part 3 of 4 - Analyze In the figure, take location O at the top surface and location at the exiting stream. Assuming the top is open to the atmosphere, then P, = Po- At location 0, by Newton's third law, the water must push on the air just as strongly as the air pushes on the water, so P, = Po- (a) By Bernoulli's law, we have Here,A, » Ag, s0 V, « V3. Assuming v, = 0 and P, = P2 = Por gives the following. V2 =v 2g(y, - Y2) 2 (9.80 m/s?)(16.3 16.3 m - 17.874 17.9 m/s Part 4 of 4- Analyze We identify the measured flow rate as AgVz=27 x 10 m/min. The formula for flow rate is as follows. (na² and solving for the diameter, we have 2.7 | × 10-³ m³/min) ( 2.7 v × 10-³ m³/min) -0.0566 X 17.9 m/s) (60 s/1 min Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. x 103 m = mm.

University Physics Volume 1

18th Edition

ISBN:9781938168277

Author:William Moebs, Samuel J. Ling, Jeff Sanny

Publisher:William Moebs, Samuel J. Ling, Jeff Sanny

Chapter14: Fluid Mechanics

Section: Chapter Questions

Problem 84P: Suppose you have a wind speed gauge like the pitot tube shown in Figure 14.32. By what factor must...

See similar textbooks

Similar questions

Explain why the fluid reaches equal levels on either side of a manometer if both sides are open to the atmosphere, even if the tubes are of different diameters.
Fluid originally flows through a tube at a rate of 100 cm3/s. To illustrate the sensitivity of flow rate to various factors, calculate be new flow rate for following changes with all other factors remaining the same as in original conditions. (a) Pressure difference increases by a factor of 1.50. (b) A new fluid wit 3.00 times greater viscosity is substituted. (c) The tube is replaced by one having 4.00 times the length. (d) Another tube used with a 0.100 times the original. (e) Yet another tube is substituted with a radius 0.100 times the original and half length, and pressure difference is increased by factor of 1.50.
Is there a limit to the height to which an entrainment device can raise a fluid? Explain your answer.
(a) Verify that a 19.0% decrease in laminar flow through a tube is caused by a 5.00% decrease in radius, assuming that all other factors remain constant. (b) What increase in flow is obtained from a 5.00% increase in radius, again assuming all other factors remain constant?
Water is moving at a velocity of 2.00 m/s through a hose with internal diameter of 1.60 cm. (a) What is the flow rate in liters per second? (b) The fluid velocity in this hose's nozzle is 15.0 m/s. What is the nozzle's inside diameter?
Do fluids exert buoyant forces in a “weightless" environment, such as in the space shuttle? Explain your answer.
What is the pressure drop due to the Bernoulli Effect as water goes into a 3.00-cm-diameter nozzle from a 9.00-cm-diameter fire hose while carrying a flow of 40.0 L/s? (b) To what maximum above be nozzle can this water rise? (The actual height will be significantly smaller due to air resistance.)
Logs sometimes float vertically a lake because one end has become water-logged and denser than the other. What is the average density of a uniform-diameter log that floats with 20.0% of its length above water?
In an immersion measurement of a woman's density, she is found to have a mass of 62.0 kg in air an apparent mass of 0.0850 kg completely submerged with lungs empty. (a) What of water does she displace? (b) What is her volume? (c) Calculate her density. (d) If her lung capacity is 1.7S L, is she able to that without treading water with her lungs filled air?
What is density of a woman floats in fresh water with 4.00% of her volume above the surface? (This could be measured by placing her in a Wii marks on the side to measure how much water she displaces when floating and when held under water.) (b) What percent of her volume is above surface when she floats in seawater?
Prove the sped of an incompressible fluid through a constriction, such as a Venturi tube, increases by a factor equal to the square of the factor by which the diameter decreases. (The converse applies for flow out of a constriction into a larger-diameter region.)
(a) Convert normal blood pressure readings of 120 over 80 mm Hg to newtons per meter squared using be relationship for pressure due to the weight of a fluid (p=hg) rater a conversion factor. (b) Explain why be blood pressure of an infant would likely be smaller than that of an adult. Specifically, consider the smaller height to which blood mast be pumped.