Suppose you pour water into a container until it reaches a depth of 20 cm. Next, you carefully pour in a 9.2 cm thickness of olive oil so that it floats on top of the water. Part A What is the pressure at the bottom of the container? Express your answer using two significant figures. |να ΑΣφ ? P = Pa

Suppose you pour water into a container until it reaches a depth of 20 cm. Next, you carefully pour in a 9.2 cm thickness of olive oil so that it floats on top of the water. Part A What is the pressure at the bottom of the container? Express your answer using two significant figures. |να ΑΣφ ? P = Pa

Name: Suppose you pour water into a container until it reaches a depth of 2
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Description: Suppose you pour water into a container until it reaches a depth of 20cm. Next, you carefully pour in a 9.2 cm thickness of olive oil so thatit floats on top of the water.Part AWhat is the pressure at the bottom of the container?Express your answer

College Physics

1st Edition

ISBN:9781938168000

Author:Paul Peter Urone, Roger Hinrichs

Publisher:Paul Peter Urone, Roger Hinrichs

Chapter12: Fluid Dynamics And Its Biological And Medical Applications

Section: Chapter Questions

Problem 61PE: Unreasonable Results A fairly large garden hose has an internal radius of 0.600 cm and a length of...

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A fluid flows through a horizontal pipe that widens, making a 45 angle with the y axis (Fig. P15.48). The thin part of the pipe has radius R, and the fluids speed in the thin part of the pipe is v0. The origin of the coordinate system is at the point where the pipe begins to widen. The pipes cross section is circular. a. Find an expression for the speed v(x) of the fluid as a function of position for x 0 b. Plot your result: v(x) versus x. FIGURE P15.48 (a) The continuity equation (Eq. 15.21) relates the cross-sectional area to the speed of the fluid traveling through the pipe. A0v0 = A(x)v(x) v(x)=A0v0A(x) The cross sectional area is the area of a circle whose radius is y(x). The widening pan of the pipe is a straight line with slope of 1 and intercept y(0) = R. y(x) = mx + b = x + R A(x) = [y(x)]2 = (x + R)2 Plug this into the formula for the velocity. Plug this into the formula for the velocity. v(x)=A0v0(x+R)2
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