QUIZ 8   Chapters 14 


Below are problems from Chapter 14  for Thursday's  (10-28) Quiz 8 , which could also include material from your corrected Quiz 6 and  (See the link "Q6" and "Q7" on the syllabus page.) . We will spend a little time reviewing the homework and sample quiz  problems before the quiz is administered. I assume you have been reading CH. 14 so by now doing these problems should be natural. Note,  homework problems from Chapter  13 are posted and  the sample quiz should  give you a fresh head  start for Q7 review.  NO LAB EXPERIMENT  THIS THURSDAY( 10-28); WE WILL USE IT TO FORGE AHEAD AND REVIEW.

RANKING : 2, 3
EXERCISES: 9, 11, 13, 16, 17, 22, 26, 29, 31, 35, 39, 40,
4. Look at the situation when the cup of water is upside down. The net force on the card must be zero; therefore the downward weight of the water on the card must  be balanced by an upward force on the card. What is the force acting upward ? Hint:  Think  about the atmosphere. 
Look at the situation when the cup of water is sideways.  The net force on the card must be zero; therefore the leftward force from the water pressure   on the card must  be balanced by a rightward force on the card. What is the force acting rightward ? Hint:  Think  about the atmosphere. 
7.  Same concepts; the net force on the water column is zero. The downward force of the weight of the water column must be balanced by an upward force.   What is the force acting upward ? Hint:  Think  about the atmosphere. 
EXERCISES: 9, 11, 13, 16, 17, 22, 26, 29, 31, 35, 39, 40,
9.  As the water bubble rises , the water pressure around it decreases. The outside force "squeezing"the bubble is less and the bubble increases in size. The air pressure on the inside of a bubble tends to equalize to the pressure in the surrounding fluid.  The bubble of course rises because of a kind  buoyant force  like a hot air balloon rising in the air. But as it rises, the water pressure goes down, and thus, the air pressure in the bubble also goes down. From Boyle's Law on page 253, the  bubble volume will increase, since p*V = P*v, where P is the larger pressure, p the smaller one, V the larger volume and v the smaller volume.
11.  The bag is like a bubble; as the pressure on the outside of the bag goes down, the outside  force "squeezing" the bag  is less. Now, the pressure inside the air tight bag does not change. But the outside pressure decreases so there is a net force pushing the bag outward to larger volumes. This  outward net  pressure is balanced  by "spring" force in the elastic bag that points toward the inside of the bag and keeps the bag  from bursting.
13. Air is compressible, unlike water and the cement making up the bricks. Thus the "squeezing" mechanism discussed in the above two problems is  in full effect. The foam rubber would be squeezed to smaller volumes, and foam bricks near the bottom of the stack will be thinner. But since they have the same mass as each of the bricks above, the density of the bottom  bricks would increase. That is what happens with air  and explains why the density of air goes down as you rise in the atmosphere: the "air bricks" higher up are less squeezed than the air bricks near the earth's surface where we reside. 
16.  The weight of the column being supported by the same air  pressure would be the same. However, since the density of the liquid  has decreased, would the column height being supported be higher or smaller. See page figure 14.7  and also think about quiz 7 which had a problem about wood and metal of equal  mass but different volumes.
17. The upward water force  holding the column up is (liquid pressure)* area. (Note that laterally, the upward [pressure from the water is equal to atmospheric pressure since the container is open to the atmosphere at the same level as the bottom of the column, the liquid pressure at the column equals atmospheric pressure. Note:
 (Atmospheric pressure)*(area of column ) = weight of column.   See figure 14.7. The downward weight of the column is (liquid mass density)*g*area*h, where h is the column's height. Thus the area cancels out when you set these two forces equal to each other. 
22. Say yes, since the large elephant displaces much more air than the small helium filled balloon. So the upward buoyant  force on the elephant is much  larger. Then why does only the  balloon rise while the elephant does not rise? 
26. More class discussion needed here.
29.  Compare the densities of air and helium. When the air is pumped out, you would have a vacuum inside, with no molecules of anything at all, either air or helium; nothing.
31.  See #22. The steel tank is like the elephant in the sense that  though the volume is the same as the balloon's, the mass is much larger. Thus, in the case of the steel container, a larger downward force prevents the tank from rising even though it has the same buoyant force as the same-sized balloon. What is the name of that  larger downward  force in the case of the steel container or any thing for that matter?
35.  See exercises #9 and #11.
39.  In the case of the non-rigid balloon, the volume will increase as the balloon  rises . It rises due to the buoyant  force and as it rises the density and pressure of the surrounding  air (outside the balloon) also will decrease, so the volume will increase even further as described in exercise  #9.   This will influence the upward buoyant  force because  more surrounding air will be displaced.  In the rigid case, the volume does not change as the balloon rises toward lower pressures.  So the buoyant force may change in a different manner than the rigid balloon case since less air seems to be displaced at the same altitude level. But think about it. 
40. The net force on the window is zero; on the outside you have a force pushing inward = (outside pressure)* (area of window) and you also have force pushing in the opposite direction to cancel out the inward directed force. These oppositely directed   forces come from the pressure inside the house and also mechanical forces securing the window in the frame.