PHY2020
Fall 2011, Test 3
I understand that the University of Florida expects its
students to be honest in all their academic work. I agree to
adhere to this commitment to academic honesty and
understand that my failure to comply with this commitment
may result in disciplinary action up to and including
expulsion from the University.
Name: _____________________________________________________
UF ID: _____________________________________________________
Ignore air friction in all problems.
Please give complete responses to all questions
including units and theoretical justification for
responses.
You must show ALL of your work to receive full credit!
Useful values and formulas:
1 km = 0.62 miles = 3280.8 feet
1 m = 39.37 inches
1 kg = 2.2 lbs
101 kPa = 760 mg Hg
ρwater = 1 g/cm3 = 1000 kg/m3
1 Cal = 4.18 Joules
cwater = 1.0 kcal/kg °C
1. (10 pts) A train emits a sound at 1645 Hz. The train is moving towards an
observer with a speed of 150 km/hr. If the observer hears a frequency of 1400
Hz, what is the speed and direction of the observer?
f = f0 •
v + vr
v + vs
1400 Hz = 1645 Hz *
(343 m/s + vr)
343 m/s – (150 km/hr * 1 hr/3600 s * 1000 m/1km)
1400 Hz = 1645 Hz * (343 + vr) / (343 – 41.67)
301.33 m/s * 1400 Hz = 1645 Hz * (343 + vr)
(421,862 m/s Hz) / 1645 Hz = 343 + vr
256.45 m/s = 343 m/s + vr
vr = -86.55 m/s
You subtract vr when if the observer is moving away from the source.
Therefore, the observer is moving 86.55 m/s, away from the train.
2. (15 pts) A spring is hanging at rest. You add a 50 kg mass to the end of the
spring and it stretches 10 cm. What is the spring constant? What is the period of
resulting oscillation? What is the frequency of oscillation?
5 pts: F = k * x
F=m*g
(50 kg) * (9.8 m/s2) = k * (10 cm * 1 m / 100 cm)
490 N = k * 0.1 m
k = 4900 kg/ s2
5 pts: T = 2π * √(m/k)
T = 2π * √(50 kg / 4900 kg/ s2)
T = 0.63 seconds
5 pts: f = 1 / T
f = 1.59 Hz
3. (10 pts) 0.5 g of liquid mercury is contained in a 5 inch tall thermometer whose
volume is 55 cm3. How much pressure is present at the bottom of the
thermometer due to the mercury?
P=ρ*g*h
5 pts: ρ = m / V, so the density of liquid mercury is 0.5 g / 1 cm3, or
(0.5 g * 1 kg/1000 g) / (55 cm3 * (1 m / 100 cm)3)
0.0005 kg / .000055 m3, or 9.10 kg/m3
5 pts: P = 9.1 kg/m3 * 9.8 m/s2 * (5 in * 1 m / 39.37 in)
P = 11.3 Pascals
4. (15 pts) A circuit has 4 resistors connected in series: one 500 Ω, one 975 Ω,
one 10 kΩ, and one 12 kΩ. If each resistor is to have 10 mA of current flowing
through it, what is the voltage needed to power this circuit?
5 pts: Rtot = (500 + 975 + 10 x 103 + 12 x 103) = 23.475 kΩ
Vtot = Itot * Rtot
5 pts: Itot = 4 * (10 * 10-3 A), = 40 * 10-3 A
5 pts” Vtot = (40 * 10-3 A) * (23475 Ω), = 939 V
5. (10 pts) 10 kg of water is cooled from 393 K to 298 K. How much heat in
Joules is required in this process?
Q = c * m * (Tf – Ti), T is in Celsius
298 K – 393 K = 24.85 °C – 119.85 °C, or -95.0 °C
Q = (1.0 kcal / kg °C) * 10 kg * (-95.0 °C)
Q = - 950 kcal
1 cal = 4.18 Joules
- 950 kcal * (1000 cal / 1 kcal) * (4.18 Joules / 1 cal) = -3.97 * 106 J
6. (15 pts) What is the speed of sound at 82 °F, and when it is freezing (32 °F)?
Explain your result using equations, ideas, and theories learned since Exam 2.
v = 331 m/s + (0.6 m/s * T), where T is in Celsius.
6 pts:
°C = (°F – 32) * 5/9
82 °F = 27.78 °C
32 °F = 0 °C
6 pts:
v = 331 + (0.6 * 27.78) = 347.67 m/s
v = 331 + (0.6 * 0) = 331 m/s
3 pts:
Sound is a mechanical wave, and it requires a medium to travel through/to
propagate the wave. As the temperature drops the molecules in the air
move more slowly. Slower moving molecules collide with other molecules,
thus transferring energy, more slowly. Thus, sound is transmitted more
slowly.
7. (15 pts) Find the electric force felt by the right electron.
Force at the point indicated is the sum of all individual electric forces.
Ftot = Fproton, right electron + Fleft electron, right electron
Fproton, right electron = (k * +e * -e) / (231 pm)2
Fleft electron = (k * -e * -e) / (200 pm)2
Where k = 9 * 109 N m2/C2 and e = 1.6 * 10-19 C
Fproton = (9 * 109 N m2/C2 * +1.6 * 10-19 C * -1.6 * 10-19 C) / (231 * 10-12 m)2
5 pts: Fproton = -4.31 * 10-9 N
Fleft electron = (9 * 109 N m2/C2 * -1.6 * 10-19 C * -1.6 * 10-19 C) / (200 * 10-12 m)2
5 pts: Fleft electron = 5.76 * 10-9 N
Ftot = -4.31 * 10-9 N + 5.76 * 10-9 N
5 pts: Ftot = 1.45 * 10-9 N of repulsion, pointing right
8. (10 pts) What electric field is felt 10 cm from a -35 pC charge? In what
direction is it pointing?
E = (k * -35 * 10-12 C) / (10 * 10-2 m)2
7 pts: E = (9 * 109 N m2/C2 * -35 * 10-12 C) / (10 * 10-2 m)2
3 pts: E = -31.50 N/C, pointing toward the charge b/c the charge is negative