f5ch15

# f5ch15 - 416 CHAPTER 15 FLUIDS CHAPTER 15 Answer to...

This preview shows pages 1–4. Sign up to view the full content.

416 CHAPTER 15 FLUIDS C HAPTER 15 Answer to Checkpoint Questions . all tie . ( a ) all tie; ( b ) : , , : . cm /s, outward . ( a ) all tie; ( b ) , then and tie, ; ( c ) , , , Answer to Questions . e, then b and d tie, then a and c tie . ( a ) ; ( b ) , less; , equal; , more . ( a ) , , ; ( b ) all tie; ( c ) no (you must consider the weight exerted on the scale via the walls) . ( a ) moves downward; ( b ) moves downward . , , , . all tie . ( a ) downward; ( b ) downward; ( c ) same . all tie . ( a ) same; ( b ) same; ( c ) lower; ( d ) higher . ( a ) increases; ( b ) increases . ( a ) block , counterclockwise; block , clockwise; ( b ) block , tip more; block , right itself

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
CHAPTER 15 FLUIDS 417 E : g = cm = ( : g = cm )( kg = g)( cm = m ) = : kg = m : E ( a ) The weight of the rst liquid is W = m g = V g = ( : g/cm )( : L)( cm = L)( cm/s ) = : dyn = : N : Similarly, for the second and the third liquid W = m g = V g = ( : g/cm )( : L)( cm = L)( cm/s ) = : N and W = m g = V g = ( : g/cm )( : L)( cm = L)( cm/s ) = : N : The total force on the bottom of the container is thus F = W + W + W = N : E The pressure increase is the applied force divided by the area: p = F=A = F= r , where r is the radius of the piston. Thus p = ( N) = [ ( : m) ] = : Pa. Notice that the value for the radius was converted to meters. E ( a ) P = lb/in. = ( lb/in. ) : Pa : lb/in = kPa : ( b ) mmHg = ( mmHg) : Pa mmHg = : kPa ; mmHg = ( mmHg) : Pa mmHg = : kPa : E The air inside pushes outward with a force given by p i A , where p i is the pressure inside the room and A is the area of the window. Similarly, the air on the outside pushes inward with a force given by p o A , where p o is the pressure outside. The magnitude of the net
R θ 418 CHAPTER 15 FLUIDS force is F = ( p i p o ) A . Since atm = : Pa, F = ( : atm : atm)( : Pa/atm)( : m)( : m) = : N. E Let the volume of the expanded air sacs be V a and that of the sh with its air sacs collapsed be V . Then sh = m sh =V = : g/cm , and w = m sh = ( V + V a ) = : g/cm . Thus ( V + V a ) =V = : = : , which gives V a = ( V + V a ) = : %. P The magnitude F of the force required to pull the lid o is F = ( p o p i ) A , where p o is the pressure outside the box, p i is the pressure inside, and A is the area of the lid. This gives p i = ( p o A F ) =A = h ( lb/in. )( in. ) lb i = ( in. ) = : lb/in. . Notice that p o is given in lb/in. and A is given in in. , so no unit conversions are required. P ( a ) At every point on the surface there is a net in- ward force, normal to the surface, due to the di er- ence in pressure between the air inside and outside the sphere. The diagram to the right shows half the sphere and some of the force vectors. We suppose a team of horses are pulling to the right. To pull the sphere apart it must exert a force at least as great as the horizontal component of the net force of the air. Consider the force acting at the angle

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern