Pre-Lab 3 Report
Adithya Sudhan
ECE 100-L02
Teammates: Peter
Prof. Oruklu
Lab Date: 9/09/15
TA: Sunny Lin
Due Date: 9/16/15
Problem Statement
To make the robot navigate out of the maze smoothly and fast.
Investigation/Research
The motor function [ for exa
Post-Lab 2 Report
Adithya Sudhan
ECE 100-L02
Teammates: Peter
Prof. Oruklu
Lab Date: 9/09/15
TA: Sunny Lin
Due Date: 9/16/15
Problem Statement
To design the robot so that it navigates out of the maze in the least time possible.
Investigation/Research
One
Pre-Lab 4 Report
Adithya Sudhan
ECE 100-L02
Teammates: Peter
Prof. Oruklu
Lab Date: 9/23/15
TA: Sunny Lin
Due Date: 9/30/15
Problem Statement
To develop a tape path following robot that uses light sensors for navigation.
Investigation/Research
The robot w
41. (a) The difference between the total amounts in freight and displacement tons, (8 7)(73) = 73 barrels bulk, represents the extra M&Ms that are shipped. Using the conversions in the problem, this is equivalent to (73)(0.1415)(28.378) = 293 U.S. b
40. (a) In atomic mass units, the mass of one molecule is 16 + 1 + 1 = 18 u. Using Eq. 1 9, we find 18u = (18u ) 1.6605402 1027 kg 1u = 3.0 1026 kg.
(b) We divide the total mass by the mass of each molecule and obtain the (approximate) number of w
37. Using the (exact) conversion 2.54 cm = 1 in. we find that 1 ft = (12)(2.54)/100 = 0.3048 m (which also can be found in Appendix D). The volume of a cord of wood is 8 4 4 = 128 ft3, which we convert (multiplying by 0.30483) to 3.6 m3. Therefore,
25. The first two conversions are easy enough that a formal conversion is not especially called for, but in the interest of practice makes perfect we go ahead and proceed formally: (a) 11 tuffets = (11 tuffets ) (b) 11 tuffets = (11 tuffets ) (c) 11
26. If we estimate the typical large domestic cat mass as 10 kg, and the typical atom (in the cat) as 10 u 2 1026 kg, then there are roughly (10 kg)/( 2 1026 kg) 5 1026 atoms. This is close to being a factor of a thousand greater than Avogradros
23. We introduce the notion of density, = m / V , and convert to SI units: 1000 g = 1 kg, and 100 cm = 1 m. (a) The density of a sample of iron is therefore
= ( 7.87 g cm
3
)
1 kg 1000 g
100 cm 1m
3
which yields = 7870 kg/m3. If we ignore
21. If ME is the mass of Earth, m is the average mass of an atom in Earth, and N is the number of atoms, then ME = Nm or N = ME/m. We convert mass m to kilograms using Appendix D (1 u = 1.661 1027 kg). Thus,
ME 5.98 1024 kg N= = = 9.0 1049 . 27 m
20. To organize the calculation, we introduce the notion of density:
=
m . V
(a) We take the volume of the leaf to be its area A multiplied by its thickness z. With density = 19.32 g/cm3 and mass m = 27.63 g, the volume of the leaf is found to be
19. We introduce the notion of density:
=
and convert to SI units: 1 g = 1 103 kg.
m V
(a) For volume conversion, we find 1 cm3 = (1 102m)3 = 1 106m3. Thus, the density in kg/m3 is 1g 1 g cm = cm 3
3
103 kg g
cm3 = 1 103 kg m 3 . 6 3 10 m
T
18. The last day of the 20 centuries is longer than the first day by
( 20 century ) ( 0.001 s
century ) = 0.02 s.
The average day during the 20 centuries is (0 + 0.02)/2 = 0.01 s longer than the first day. Since the increase occurs uniformly, the
38. (a) Squaring the relation 1 ken = 1.97 m, and setting up the ratio, we obtain
1 ken 2 1.97 2 m 2 = = 3.88. 1 m2 1 m2
(b) Similarly, we find 1 ken 3 197 3 m3 . = = 7.65. 3 1m 1 m3 (c) The volume of a cylinder is the circular area of its base mult
39. (a) For the minimum (43 cm) case, 9 cubit converts as follows: 9cubit = ( 9cubit ) And for the maximum (43 cm) case we obtain 9cubit = ( 9cubit ) 0.53m = 4.8m. 1cubit 0.43m = 3.9m. 1cubit
(b) Similarly, with 0.43 m 430 mm and 0.53 m 530 mm, we
Post-Lab 8 Report
ECE 100-01
Prof. Oruklu
TA: Sijia Wu
Alexander Gooden
Isaac Gewarges
Nasif Rabbi
Lab Date: 11/06/14
Due Date: 11/10/14
Problem Statement
To build a robot that can navigate a taped path to play a game named Mint
Shuffle X in which the rob
Pre-Lab 2 Report
Adithya Sudhan
ECE 100-L 02
Teammates: Peter
Prof. Oruklu
Lab Date: 9/02/15
TA: Sunny Lin
Due Date: 9/9/15
Problem Statement
To incorporate metasensing into the code and improve navigation.
Investigation/Research
The textbook offers a cle
Post-Lab 1 Report
Adithya Sudhan
ECE 100-L 02
Teammates: Peter
Prof. Oruklu
Lab Date: 9/01/15
TA: Sunny Lin
Due Date: 9/08/15
Problem Statement
To analyse the results of building the Handy Bug.
Investigation/Research
The goal of the lab session was to bui
44. According to Appendix D, a nautical mile is 1.852 km, so 24.5 nautical miles would be 45.374 km. Also, according to Appendix D, a mile is 1.609 km, so 24.5 miles is 39.4205 km. The difference is 5.95 km.
45. (a) The receptacle is a volume of (40)(40)(30) = 48000 cm3 = 48 L = (48)(16)/11.356 = 67.63 standard bottles, which is a little more than 3 nebuchadnezzars (the largest bottle indicated). The remainder, 7.63 standard bottles, is just a little les
48. The volume of the water that fell is
V = ( 26 km
) ( 2.0 in.) = ( 26 km ) = ( 26 10 m ) ( 0.0508 m )
2 2 6 2
1000 m 1 km
2
( 2.0 in.)
0.0254 m 1 in.
= 1.3 106 m 3 .
We write the mass-per-unit-volume (density) of the water as:
=
m = 1
46. The volume of the filled container is 24000 cm3 = 24 liters, which (using the conversion given in the problem) is equivalent to 50.7 pints (U.S). The expected number is therefore in the range from 1317 to 1927 Atlantic oysters. Instead, the numbe
50. The volume of one unit is 1 cm3 = 1 106 m3, so the volume of a mole of them is 6.02 1023 cm3 = 6.02 1017 m3. The cube root of this number gives the edge length: 8.4 105 m3 . This is equivalent to roughly 8 102 kilometers.
47. We convert meters to astronomical units, and seconds to minutes, using
1000 m = 1 km 1 AU = 1.50 108 km 60 s = 1 min . Thus, 3.0 108 m/s becomes 3.0 108 m s 1 km 1000 m AU 1.50 108 km 60 s = 0.12 AU min . min
49. Equation 1-9 gives (to very high precision!) the conversion from atomic mass units to kilograms. Since this problem deals with the ratio of total mass (1.0 kg) divided by the mass of one atom (1.0 u, but converted to kilograms), then the computat
51. A million milligrams comprise a kilogram, so 2.3 kg/week is 2.3 106 mg/week. Figuring 7 days a week, 24 hours per day, 3600 second per hour, we find 604800 seconds are equivalent to one week. Thus, (2.3 106)/(604800) = 3.8 mg/s.
17. The time on any of these clocks is a straight-line function of that on another, with slopes 1 and y-intercepts 0. From the data in the figure we deduce
2 594 tB + 7 7 33 662 tB = tA . 40 5 tC =
These are used in obtaining the following result