INTRODUCTORY PHYSICS  TEST NO. 3  SPRING 2007

     1. (a) An alveolus in the foam model is in equilibrium during
	exhalation with a pressure of 8.0 torr and a radius of 0.10 mm.
	What is the surface tension of the fluid coating the alveolus?
	(b) During inhalation the pressure increases to 18.0 torr and the
	radius to 0.18 mm.  What must the surface tension of the coating
	fluid become to maintain equilibrium?  Note that 1 torr equals
	1333 dynes per square centimeter.

	In the foam model, only one layer of surfactant is present and
	the surface tension is given by Pr/2.  Therefore

		ST = (8.0 torr)(1333 dyn/cm2/torr)(0.01 cm) / 2 = 53 dyn/cm

	For equilibrium to be maintained in the new situation, the surface
	tension must change according to ST = Pr/2 to meet the new
	conditions.

		ST = (18 torr)(1333 dyn/cm2/torr)(0.018 cm) / 2 = 216 dyn/cm

     2. A cylinder of metal with a mass of 50 g and a temperature of 90 C
	is placed in water with a mass of 50 g and a temperature of 10 C.
	The final temperature of the metal plus water (when they are in
	thermal equilibrium) is 15 C.  Assuming no heat lost to or gained
	from the environment, what is the specific heat of the metal?
	Recall that the specific heat of water is 1 cal/g/degree C.

	The equation for specific heat is

		Q = ms(T2 - T1).

	First calculate the heat gained by the water from the metal.  This
	is

		Q = (50 g)(1 cal/g-°C)(15°C - 10°C) = 250 cal

	The heat lost by the metal involves the unknown specific heat of
	the metal, smetal.  The heat lost by the metal must
	be the 250 cal gained by the water.

		250 cal = (50 g)smetal(90°C - 15°C)

	Solving for smetal

		smetal = (250 cal)/[(50 g)(75°)] = 0.067 cal/g·°C

     3. A long steel measuring tape is 100 yards long at a temperature of
	30 C.  Just before a football game, the tape is to be used to mark
	off the field.  If the temperature is 0 C, how much shorter will
	the tape be?  Express your answer in the units of inches.  (1 yard
	= 3 feet, 1 foot = 12 in.) The coefficient of linear expansion for
	steel is 0.000013 per degree C.

	The equation to use for linear thermal expansion is

		ΔL = LoαΔT

	We are after the change in L, ΔL, in this equation.

		ΔL = (100 yd)(3 ft/yd)(12 in/ft)(0.000013 °C-1)(0 - 35°C) = -1.4 in

	Therefore the tape will be 1.4 inch shorter.

     4. Two charges of 8.5 microcoulombs (millionth of a coulomb) each
	repel each other with a force of 10.0 millinewtons (thousandth of
	a newton).  How far apart are the charges?

	The coulomb force is calculated from

		F = k q1q2/r2.

	For this problem

		10 × 10-3 N = (9 × 109 N·-m2/C2)(8.5 × 10-6 C)2 / r2

	or

		r2(10 × 10-3 N) = (9 × 109 N·m2/C2)(8.5 × 10-6 C)2

	or

		r2 = (9 × 109 N·m2/C2)(8.5 × 10-6 C)2/(10 × 10-3 N)

		r = 8.1 m

     5. A resistor of 6 ohms and one of 18 ohms are placed in parallel
	across  a 12 V battery.  (a) What is the current flowing in each
	resistor?  (b) What is the power being dissipated by each resistor?
	(c) What is the total current and power in the circuit?

     (a)

	I1 = 12 V / 6 ohms = 2.0 A

	I2 = 12 V / 18 ohms = 0.67 A

     (b)

	P1  = I1V = (2.0 A)(12 V) = 24 W

	P2  = I2V = (0.67 A)(12 V) = 8.0 W

     (c)

	I = I1 + I2 = 6.67 A

	P = P1 + P2 = 32 W