ch14 - 1 The air inside pushes outward with a force given...

Info iconThis preview shows pages 1–8. Sign up to view the full content.

View Full Document Right Arrow Icon
1. 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 force is F = ( p i p o ) A . Since 1 atm = 1.013 × 10 5 Pa, 54 (1.0 atm 0.96 atm)(1.013 10 Pa/atm)(3.4 m)(2.1 m) = 2.9 10 N. F =− × ×
Background image of page 1

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

View Full Document Right Arrow Icon
2. We note that the container is cylindrical, the important aspect of this being that it has a uniform cross-section (as viewed from above); this allows us to relate the pressure at the bottom simply to the total weight of the liquids. Using the fact that 1L = 1000 cm 3 , we find the weight of the first liquid to be 11 1 1 33 2 6 2 (2.6 g / cm )(0.50 L)(1000 cm / L)(980 cm/ s) 1.27 10 g cm/s 12.7 N. Wm g V g ρ == × = In the last step, we have converted grams to kilograms and centimeters to meters. Similarly, for the second and the third liquids, we have 2 22 2 2 (1.0 g/cm )(0.25 L)(1000 cm L)(980 cm s ) 2.5 N g V g = = and 2 3 3 (0.80 g/cm )(0.40 L)(1000 cm / L)(980 cm/s ) 3.1 N. g V g = = The total force on the bottom of the container is therefore F = W 1 + W 2 + W 3 = 18 N.
Background image of page 2
3. The pressure increase is the applied force divided by the area: p = F / A = F / π r 2 , where r is the radius of the piston. Thus p = (42 N)/ π (0.011 m) 2 = 1.1 × 10 5 Pa. This is equivalent to 1.1 atm.
Background image of page 3

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

View Full Document Right Arrow Icon
4. The magnitude F of the force required to pull the lid off 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. Recalling that 1N/m 2 = 1 Pa, we obtain 54 42 480 N 1.0 10 Pa 3.8 10 Pa. 77 10 m io F pp A =−=× = × ×
Background image of page 4
5. Let the volume of the expanded air sacs be V a and that of the fish with its air sacs collapsed be V . Then 33 fish fish fish 1.08 g/cm and 1.00 g/cm w a mm VV V ρρ == = = + where ρ w is the density of the water. This implies fish V = w ( V + V a ) or ( V + V a )/ V = 1.08/1.00, which gives V a /( V + V a ) = 7.4%.
Background image of page 5

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

View Full Document Right Arrow Icon
6. Knowing the standard air pressure value in several units allows us to set up a variety of conversion factors: (a) () 5 2 2 1.01 10 Pa 28 lb/in. 190 kPa 14.7 lb/in P §· × == ¨¸ ©¹ (b) 55 1.01 10 Pa (120 mmHg) 15.9 kPa, (80 mmHg) 10.6 kPa. 760 mmHg 760 mmHg § · ×× ¨ ¸ © ¹
Background image of page 6
7. (a) The pressure difference results in forces applied as shown in the figure. We consider a team of horses pulling to the right. To pull the sphere apart, the team must exert a force at least as great as the horizontal component of the total force determined by “summing” (actually, integrating) these force vectors. We consider a force vector at angle θ . Its leftward component is p cos dA , where dA is the area element for where the force is applied. We make use of the symmetry of the problem and let dA be that of a ring of constant on the surface. The radius of the ring is r = R sin , where R is the radius of the sphere. If the angular width of the ring is d , in radians, then its width is R d and its area is dA = 2 π R 2 sin d . Thus the net horizontal component of the force of the air is given by 2 22 2 / 2 2 0 0 2 sin cos sin .
Background image of page 7

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

View Full Document Right Arrow Icon
Image of page 8
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 99

ch14 - 1 The air inside pushes outward with a force given...

This preview shows document pages 1 - 8. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online