Physics100: HW 6 Solutions
Chapter 14 and 15:
1) The pressure exerted against the ground by an elephant’s weight distributed evenly over its four feet
is less than 1 atmosphere. Why, then, would you be crushed beneath the foot of an elephant, while
you’re unharmed by the pressure of the atmosphere?
If an elephant steps on you, the pressure that the elephant exerts is over and above the atmospheric
pressure that already is exerted on you. It is the extra pressure the elephant’s foot produces that
crushes you. For example, if atmospheric pressure the size of an elephant’s foot were somehow
removed from a patch of your body, you would be in serious trouble. You would be soothed, however, if
an elephant stepped onto this area!
2) Estimate the buoyant force that air exerts on you. To do this, you can estimate your volume by
knowing your weight and by assuming your weight density is a bit less than that of water.
Let’s say my weight is 120 pounds = 534 N. I estimate my weight-density to be a bit less than that of
water, say 9500 N/m3. (water’s mass-density is 1000 kg/m3 which corresponds to a weight-density of
9800 N/m^3). So, since weight-density = weight/volume, this means my
volume = weight/weight-density = 534 N/(9500 N/m3) = 0.056 m3
The buoyant force that air exerts on me equals the weight-density of air multiplied by my volume, and
since the density of air is about 1.25 kg/m3, so weight-density of air is 9.8 times this, i.e. 12.25 N/m3, we
have,
buoyant force = my volume x weight-density-of-air = 0.056 m3 x 12.25 N/m3 = 0.69 N
i.e. the buoyant force on me is about 0.7 N.
This is much smaller than my weight!
3) Desert sand is very hot during the day and very cool at night. What does this tell you about its specific
heat?
Sand has a low specific heat, as evidenced by its relatively large temperature changes (for small changes
in internal energy), i.e. a small thermal inertia. A substance with a high specific heat such as water, on
the other hand, must absorb or give off large amounts of internal energy for comparable temperature
changes.
4) Suppose that water is used in a thermometer instead of mercury. If the temperature is at 4oC and
then changes, why can’t the thermometer indicate whether the temperature is rising or falling?
Water has the greatest density at 4°C; therefore, either cooling or heating at this temperature will result
in an expansion of the water. A small rise in water level would be ambiguous, as it could mean that
either the temperature has increased or decreased, and so this makes a water thermometer impractical
in this temperature region.