Physics 102 Lawrence Weinstein Homework Set 3 Solutions Chapter 14: Exercises: 4: The ridges are on the outside of the funnel. Liquid flows down into the bottle through the funnel. Therefore, the air in the bottle must be able to leave somehow to make room for the liquid. The ridges on the funnel provide the space for the air to leave the bottle. 7: When an air bubble rises in water, its mass stays the same (since it does not collect any more air from the water). The pressure decreases as it rises. The temperature stays the same since it is in the water and the water is presumably all at the same temperature. Since the pressure decreases, the volume must increase to compensate (since pv = constant). If the volume increases and the mass is the same, then the density decreases. 10: The air pressure at 10,000 m (30,000 feet) is much less than the air pressure at sea level. The airplanes are pressurized to increase the pressure inside the plane, but it is still much less than at sea level. Thus, there is much less pressure in the airplane so the gas in the package expands, puffing up the package. This problem is just like problem 7. 16: Air pressure pushes up on the water with a pressure of 10^5 pascals. If the pump is a perfect pump and can create a perfect vacuum, then there will be zero pressure pushing down on the water. In this case, the maximum column of water than can be lifted will be one where the pressure of the column is just barely smaller than atmospheric pressure. For water, this is a column height of 10 m. Thus, the deepest well that a pump can extract water from is 10 m (assuming the pump is at the surface). 22: You are unharmed by air pressure because you have compensating pressure inside you pushing out (like the soda can we crushed in class). When the elephant steps on you, it drastically increases the force on you without any compensating internal pressure increase. This has nothing to do with Newton's 3rd Law. 32: As anyone who has ever tried to inflate a rubber balloon knows, it takes extra force to make the rubber expand. Therefore, the rubber of the balloon exerts an inward force on the air. This force produces a pressure (since pressure = force / area) that increases the gas pressure inside the balloon. Note that this extra pressure is there for air balloons, helium balloons, etc. (If instead of a rubber balloon, you used a plastic bag full of air, then the internal and external pressures would be the same [since it does not take extra force to expand the plastic bag].) 34: Assuming that neither balloon pops, the expanding balloon will float higher. This is because the buoyant force is due to the weight of the displaced air. As the balloons rise, the air pressure decreases and hence the air density decreases. This causes the weight of the displaced air to decrease and hence the buoyant force decreases. The rigid balloon always has the same volume so it always displaces the same volume of air. The expanding balloon expands as it rises (since the pressure decreases) and thus displaces more air as it rises. Thus, at any altitude, the expanding balloon will have a greater buoyant force. 46: The wind blows rapidly over the wavetops, lowering the pressure there. This makes the wave tops increase in height. The wind does not blow rapidly in the troughs (since they are sheltered by the wave peaks), thus the pressure in the troughs does not decrease. The pressure difference between peak and trough increases the height of the waves. ---------------------------------- Problems: Chapter 14: 2: Since pressure times volume is constant (at constant temperature), if you decrease the volume by a factor of ten, you increase the pressure by a factor of ten. 4: To get the buoyant force, I need to determine my volume. To get this, I can use my mass and my density. I know that my density is very close to 1 (since I barely float in water). My mass is about 170 pounds or 77 kg. Since mass = density * volume, my volume is V = M/d = 77 kg / 1000 kg/m^3 = 0.077 m^3. The density of air (from the table on page 270) is 1.2 kg/m^3 at room temperature so the mass of the displaced air is 0.077 m^3 * 1.2 kg/m^3 = 0.093 kg. The weight of the displaced air is 0.093 kg * 10 m/s^2 = 0.93 N or a little less than 1 newton. Very little actually. ---------------------------------- Estimation: There are approximately 6 billion people = 6*10^9 people in the world today. If I assume that everyone picks their nose and that they spend 10 seconds each day in this rewarding activity, then the fraction of time that someone spends each day picking their nose is Fraction = 10 s / 1 day = 10 s / (1 day * 24 hours/day * 60 minutes/hour * 60 seconds/minute) = 10 s / 86400 s = approximately 10 / 10^5 = 10^(-4). If each person spends one 10-thousandth of his or her day picking their nose, then we can assume that one 10-thousandth of the people in the world are picking their nose at any given instant. This means that we have 6 * 10^9 people * 10^(-4) = 6*10^5 people picking their nose right at this very instant. That's more than half a million people! Eeeeyew!