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

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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
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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
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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
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This note was uploaded on 01/22/2012 for the course PHYS 2101 taught by Professor Grouptest during the Fall '07 term at LSU.

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

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