1) the Pressure Exerted Against the Ground by an Elephant S Weight Distributed Evenly Over

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, andso this makes a water thermometer impractical in this temperature region.