INTRODUCTORY PHYSICS  TEST NO. 2  SPRING 08   NAME KEY               


     PROBLEMS.  Each worth 10 points.  Show your work for full credit.  Ex-
                press answers in the proper units.

     1. A man has sunk in a swimming pool by exhaling all his breath and is
	standing on the bottom.  If the man is 1.6 m tall, what is the diffe-
	rence in pressure between the top of his head and the bottom of his
	feet in pascal?  The density of water is 1000 kg per cubic meter.

	The difference in pressure is due to the difference in the vertical
	position in the water between his head and feet.  That difference is
	1.6 m.  Therefore the pressure difference has to be

	ΔP = dgh = (1000 kg/m3)(9.8 N/kg)(1.6 m) = 1.6 × 104 N/m2 = 16 kPa.


     2. An object with a volume of 0.025 cubic meter is submerged completely
	in water.  What is the buoyant force in newtons acting on the object?
	(The density of water is 1000 kg per cubic m.)  If the object has an
	apparent weight of 40 N in the water, what is its actual weight?

	The buoyant force is the weight of the displaced water by Archimedes'
	principle. Since 0.025 cubic meter of water is displaced, the
	buoyant force is

	W = mg,

	where m is the mass of the displaced water, equal to the density of
	water, d, times volume, V.

	W = dVg = (1000 kg/m3)(0.025 m3)(9.8 N/kg) = 245 N.

	Since the object appears to weigh 40 N when submerged, and a 245 N
	buoyant force is acting upwards on it, its actual weight must be

	245 N + 40 N = 285 N.


     3. Weather balloons are designed to expand as they rise to high alti-
	tudes to accomodate the expansion of the helium gas inside the bal-
	loon.  Consider a weather balloon containing 0.200 cubic meter of
	helium at the surface where the pressure is one atmosphere and the
	temperature 17 deg C.  It rises to an altitude where the pressure is
	0.250 atmosphere and the temperature -53 deg C.  What would the
	volume of the helium be then?  (Recall that 0 deg C = 273 K.)

	This is an application of the equation

	P1V1 / T1 = P2V2 / T2,

	where P1 = 1.00 atm, P2 = 0.25 atm, T1 = 17° C, T2 = -53° C, and V1 = 0.20 m3.

	However, you must have absolute temperature in this equation, so you
	have to convert temperature units to kelvin.  Thus,

	T1 = l7° C + 273 = 290 K
	T2 = -53° C + 273 = 220 K.

	Re-arranging the equation to solve for V2 yields

	V2 = V1(T2 / T1)(P1 / P2) = (0.20 m3(220 K / 290 K)(1 atm / 0.25 atm) = 0.61 m3.


     4. Using Poiseiulle's equation, what is the resistance to flow of a
	liquid with a viscosity of 0.040 poise in a tube 0.80 cm in radius
	and 50 cm long?  If a pressure of 220 dynes per square centimeter
	were applied across the tube, what would the volume flow rate be?
	If a new tube of radius 1.00 cm were substituted for the original
	tube, all other factors remaining constant, what would the new volume
	flow rate be?  Recall: Area of a circle = pi times the radius squared.

	(Note:  The formula for the area of a circle should have been attached
	to problem 5, but at least you had it.)

	The resistance by Poiseiulle's law is

	R = 8ηL / πr4 = 8(0.04 poise)(50 cm) / π(0.8 cm)4 = 12.4 poise/cm3,

	to three significant figures.  Note that the poise is a cgs unit equal
	to dyne-s/cm2, so you could have expressed your answer as

	12.4 dyne-s/cm5,

	but this wasn't necessary.  Whichever way you express the units of the
	resistance, you know the volume flow rate will come out in cubic
	centimeters per second, because all units used are cgs.  The volume
	flow rate is

	F = ΔP / R = (220 dyne/cm2) / (12.4 poise/cm3) = 18 cm3/s.

	The flow rate scales as the radius to the fourth power, so if the
	radius changes from 0.80 cm to 1.00 cm, the flow rate will become

	(18 cm3/s)(1.00 cm / 0.80 cm)4 = 43 cm3/s.


     5. Liquid flows horizontally from a larger tube of radius 0.50 cm to a
	smaller tube of radius 0.35 cm.  The average speed of the liquid in
	the larger tube is 40 cm/s.  What is the volume flow rate?  What is
	the average speed of the liquid in the smaller tube?  Does the pres-
	sure go up or down in the smaller tube and by how much?  (Assume the
	liquid is water with a density of 1 gram per cubic centimeter.
	Express your answer in dynes per square centimeter.)

	Volume flow rate is defined as average speed times cross-sectional
	area, so

	F = vA = (40 cm/s)[π(0.5 cm)2] = 31 cm3/s.

	The volume flow rate in the smaller tube must also be 31 cm3/s,
	so just turn the equation around:

	v = F / A = (31 cm3/s) / [π(0.35 cm)2] = 82 cm/s.

	Since the flow speeds up in the smaller tube, the pressure will drop
	due to the Bernoulli effect.  Since the flow is horizontal, the
	change in pressure is given by

	ΔP = (1/2)dv22 - (1/2)dv12 = (1/2)(1 g/cm3)[(40 cm/s)2 - (82 cm/s)2] = -2500 dyne/cm2,

	a drop of 2500 dyne/cm2.