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.

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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Question

Part 3 of 4 - Analyze
In the figure, take location O at the top surface and location 2 at the exiting stream. Assuming the top is
open to the atmosphere, then P, = Po. At location 2, by Newton's third law, the water must push on the air
just as strongly as the air pushes on the water, so P2 = Po-
(a) By Bernoulli's law, we have
P」 + 글pri2+ pgy1= P2 + 극p22 + pgy2
Here, 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/s
16.3
16.3 m
= 17.874
Part 4 of 4 - Analyze
We identify the measured flow rate as
A2V2= [2.7
x 10-3 m3/min.
The formula for flow rate is as follows.
nd?
V21
and solving for the diameter, we have
1
2.7
v x 10-3 m/min)
4
d =
IT V2
2.7
x 10-3 m/min) 2
0.0566 X
m/s) (
60 s/1 min
=
17.9
Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. x 10-3
m
mm.

A large storage tank, open at the top and filled with water, develops a small hole in its side at a point 16.3 m
below 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.
1
Part 1 of 4 - Conceptualize
An object falling freely from height h from rest reaches a speed v = V 2gh. For ideal flow, we can expect this
same speed for the spurting water. A few millimeters seem reasonable for the diameter of a hole on the side
of a tank that sprays out only a couple of liters every minute.
Part 2 of 4 - Categorize
We use Bernoulli's equation and the expression for volume flow rate contained in the equation of continuity for
fluids.
Part 3 of 4 - Analyze
In the figure, take location O at the top surface and location 2 at the exiting stream. Assuming the top is
open to the atmosphere, then P, = Po. At location 2, 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
P, +pv,? + pgy, = P2 + pv² + pgy2:

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