Physics 202 Homework, Day 01: Chapter 12 : 2, 11, 19, 20 + In-Class Question
2. On the moon the surface temperature ranges from 375 K during the day to 1.00 x 102 K at night.
What are these temperatures on the (a) Celsius and (b) Fahrenheit scales ?
Celsius1 = 375. - 273
102.
Celsius2 = 100. - 273
- 173.
Fahrenheit1 = Celsius1 *
9
5
+ 32
215.6
Fahrenheit2 = Celsius2 *
9
5
+ 32
- 279.4
11. Find the approximate length of the Golden Gate Bridge if it is known that the steel in the roadbed expands by 0.53
m when the temperature changes from + 2 °Ç to + 32 °C.
In[177]:=
aSteel = 12 * 10-6 ;
DT = 32. - 2.;
DL = .53;
DL
L0 =
aSteel * DT
1472.22
Out[180]=
In[181]:=
Print@L0 * 3.28, " Feet long"D
4828.89 Feet long
This is close to the length of the span between the two towers (4200 feet), not the total bridge length (8981 feet). Using 0.46 meters would give the correct
span length, or 0.98 meters for the total length.
19. A simple Pendulum consists of a ball connected to one end of a thin brass wire.The period of the pendulum is
2.0000 s.The temperature rises by 140 C°, and the length of the wire increases.
Determine the period of the heated pendulum.
[Hint :] the pendulum formula is found in the chapter on simple harmonic motion.
f = 1/T =
1
g
2p
L
g = 9.8;
PendulumLength =
2.02
H2 pL2
* g; Print@PendulumLength, " meter"D
0.992948 meter
aBrass = 19 * 10-6 ;
DLength = PendulumLength * aBrass * 140.0;
Print@DLength, " meter -> change in pendulum length"D
0.00264124 meter -> change in pendulum length
2
Day_01_sol.nb
NewLength = PendulumLength + DLength;
period = 2 p
NewLength
g
; Print@period, " second -> new pendulum period"D
2.00266 second -> new pendulum period
LengthOfDay = 24 * 60 * 60; Print@LengthOfDay, " seconds in a day"D
OldTicks = LengthOfDay ê 2;
NewTicks = LengthOfDay ê period;
Error = HOldTicks - NewTicksL * period; Print@"The error would be ", Error, " secondsêday"D
86 400 seconds in a day
The error would be 114.836 secondsêday
20. Concrete sidewalks are always laid in sections, with gaps between each section. For example, the drawing shows
three identical 2.4 m sections, the outer two of which are against immovable walls. The two identical gaps between the
sections are provided so that thermal expansion will not created thermal stress that could lead to cracks.
What is the minimum gap width necessary to account for an increase in temperature of 32 C°?
Solution : The total of the two gaps must be equal to the change in length of the three concrete sections, so Gap = 2/3 * DL.
In[182]:=
L0 = 2.4; aConcrete = 12 * 10-6 ; DT = 32.;
DL = L0 * aConcrete * DT;
3
Gap =
DL; Print@"The gap must be ", Gap * 1000, " millimeter."D
2
The gap must be 1.3824 millimeter.
In - Class Question: A vintage carʼs gas tank holds exactly 10.0 gallons. Gas is pumped from an underground storage
tank at a temperature of 40 °F.
It is a hot summer day, and the car and the gasoline warm up to 90 °F, how much gasoline will spill out of the tank?
Assume the expansion of the gas tank can be ignored and that there is no system to prevent gasoline spills (as in
modern cars).
V0 = 10.; bGasoline = 9.6 * 10-4 ;
5
Celsius1 =
H40. - 32L;
9
Celsius2 =
5
H90. - 32L;
9
DT = Celsius2 - Celsius1; Print@"DT = ", DT, " Celsius Degrees"D
DV = V0 * bGasoline * DT;
Print@DV, " gallons, or ", DV * 16, " cups"D
DT = 27.7778 Celsius Degrees
0.266667 gallons, or 4.26667 cups
This is a significant waste of fuel and a source of air pollution (hydrocarbons).