Do NOT write on these sheets or take them with you! The next class needs them too!
PHYS-1200 PHYSICS II Class 16 Activity: Beats and the Doppler Effect Beats SOLUTIONS
SPRING 2006
In this part of the activity you will study the phenomenon of beats
Name: '- ' l 7 l Section:
PHYS-1200 PHYSICS II Fall 2013
Lab 4: Capacitance
Part 1. Applying Gauss's Law to find Capacitance
Its the year 2026. You are an engineer working for Spacely Space
Sprockets on the big Flux Capacitor project. You have a situa
NAME: Section:
D («7% oxL
PHYS-1200 PHYSICS 11 Fall 2013
Lab 1: Electrostatics --- Electric Charge, Coulombs Law and Electric Field
Important Concepts to start with. . . .
All matter contains enormous amounts of charge. Is all charge negative, such a
NAME and Section:
PHYS-1200 PHYSICS 11 Fall 2013
Lab 2: Flux and Gauss' Law
1. Area Vectors
On the gure, draw an area vector A on the object shown below for the at side facing you. On
_,
the gure, draw two different area vectors A on the curved side o
7 , p _ l- at?
Name: SCOTT U 2 46 Section:
PHYSICS 1200 Lab 7: Magnetism
1 ne objective of the rst third of Physics 11 is the development of Maxwells Equations. In integral form:
t
_ qenc
E d
£0
fB-da=0
d
fE-ds= E¢~,:1
fB-ds= [101+ souoatCDE
50. The full extent of the bolt's fall is given by y yo =
1 2
g t2 where y yo = 90 m (if
upwards is chosen as the positive y direction). Thus the time for the full fall is found to be t = 4.29 s. The first 80% of its free fall distance is given by 72 = g
17. None of the clocks advance by exactly 24 h in a 24-h period but this is not the most important criterion for judging their quality for measuring time intervals. What is important is that the clock advance by the same amount in each 24-h period. The cl
51. The speed of the boat is constant, given by vb = d/t. Here, d is the distance of the boat from the bridge when the key is dropped (12 m) and t is the time the key takes in falling. To calculate t, we put the origin of the coordinate system at the poin
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 cumulati
52. The y coordinate of Apple 1 obeys y yo1 =
1 2
g t2 where y = 0 when t = 2.0 s.
This allows us to solve for yo1, and we find yo1 = 19.6 m. The graph for the coordinate of Apple 2 (which is thrown apparently at t = 1.0 s with velocity v2) is y yo2 = v2
19. When the Sun first disappears while lying down, your line of sight to the top of the Sun is tangent to the Earth's surface at point A shown in the figure. As you stand, elevating your eyes by a height h, the line of sight to the Sun is tangent to the
53. (a) With upward chosen as the +y direction, we use Eq. 2-11 to find the initial velocity of the package: v = vo + at 0 = vo (9.8 m/s2)(2.0 s)
which leads to vo = 19.6 m/s. Now we use Eq. 2-15: y = (19.6 m/s)(2.0 s) +
1 2
(9.8 m/s2)(2.0 s)2 20 m .
We n
20. The density of gold is
=
m 19.32 g = = 19.32 g/cm3 . 3 V 1 cm
(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
V = We convert the
54. We use Eq. 2-16, vB2 = vA2 + 2a(yB yA), with a = 9.8 m/s2, yB yA = 0.40 m, and vB 1 = 3 vA. It is then straightforward to solve: vA = 3.0 m/s, approximately.
21. We introduce the notion of density:
=
and convert to SI units: 1 g = 1 10-3 kg.
m V
(a) For volume conversion, we find 1 cm3 = (1 10-2m)3 = 1 10-6m3. Thus, the density in kg/m3 is
1g 1 g cm = cm3
3
10-3 kg g
cm 3 = 1 103 kg m3 . -6 3 10 m
Thus, the ma
55. (a) We first find the velocity of the ball just before it hits the ground. During contact with the ground its average acceleration is given by aavg = v t
where v is the change in its velocity during contact with the ground and t = 20.0 10-3 s is the d
22. (a) We find the volume in cubic centimeters 193 gal = (193 gal ) 231 in 3 1 gal 2.54 cm 1in
3
= 7.31 105 cm3
and subtract this from 1 106 cm3 to obtain 2.69 105 cm3. The conversion gal in3 is given in Appendix D (immediately below the table of Volume
56. (a) We neglect air resistance, which justifies setting a = g = 9.8 m/s2 (taking down as the y direction) for the duration of the motion. We are allowed to use Eq. 2-15 (with y replacing x) because this is constant acceleration motion. We use primed va
16. Since a change of longitude equal to 360 corresponds to a 24 hour change, then one expects to change longitude by 360 / 24 = 15 before resetting one's watch by 1.0 h.
49. We neglect air resistance, which justifies setting a = g = 9.8 m/s2 (taking down as the y direction) for the duration of the motion. We are allowed to use Table 2-1 (with y replacing x) because this is constant acceleration motion. We are placing the
15. 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 tC = 2 594 , tB + 7 7 tB = 33 662 . tA - 40 5
These are used in obtaining the following results. (a
8. We make use of Table 1-6. (a) We look at the first ("cahiz") column: 1 fanega is equivalent to what amount of cahiz? We note from the already completed part of the table that 1 cahiz equals a dozen fanega. 1 Thus, 1 fanega = 12 cahiz, or 8.33 10-2 cahi
41. The positions of the cars as a function of time are given by
1 1 xr (t ) = xr 0 + ar t 2 = (-35.0 m) + ar t 2 2 2 xg (t ) = xg 0 + vg t = (270 m) - (20 m/s)t
where we have substituted the velocity and not the speed for the green car. The two cars pass
42. In this solution we elect to wait until the last step to convert to SI units. Constant acceleration is indicated, so use of Table 2-1 is permitted. We start with Eq. 2-17 and denote the train's initial velocity as vt and the locomotive's velocity as v
9. We use the conversion factors found in Appendix D. 1 acre ft = (43,560 ft 2 ) ft = 43,560 ft 3 Since 2 in. = (1/6) ft, the volume of water that fell during the storm is
V = (26 km 2 )(1/6 ft) = (26 km 2 )(3281ft/km) 2 (1/6 ft) = 4.66 107 ft 3 .
Thus,
V
43. (a) Note that 110 km/h is equivalent to 30.56 m/s. During a two second interval, you travel 61.11 m. The decelerating police car travels (using Eq. 2-15) 51.11 m. In light of the fact that the initial "gap" between cars was 25 m, this means the gap ha
10. A day is equivalent to 86400 seconds and a meter is equivalent to a million micrometers, so 3.7 m 106 m m = 31 m s . . 14 day 86400 s day
b
b
gc gb
h
g
44. Neglect of air resistance justifies setting a = g = 9.8 m/s2 (where down is our y direction) for the duration of the fall. This is constant acceleration motion, and we may use Table 2-1 (with y replacing x). (a) Using Eq. 2-16 and taking the negative
11. A week is 7 days, each of which has 24 hours, and an hour is equivalent to 3600 seconds. Thus, two weeks (a fortnight) is 1209600 s. By definition of the micro prefix, this is roughly 1.21 1012 s.
PHYS-1200 PHYSICS II
EXAM 1A
Sept 26, 2007
NAME _
Exams for each section are given in the rooms listed. If you are in the wrong section, move to
the right one immediately.
SECTION
EXAMINATION
NUMBER
ROOM
1
DCC 308
2
DCC 308
3
SA 3303
4
SA 3303
5
DCC 318
6
Topics for Exam 1
Chapter and Topic
Ch 21 Electric Charge and
Coulombs Law
Detailed Topics
Sections 1, 2, 3, 4, 5, 6
Nature of electric charges and forces
Vector forces
Coulombs Law
Ch 22 Electric Fields
Sections 2,3,4,5,6,7,8 (not 9)
Definition of field
PHYS-1200 PHYSICS II
EXAM 1A
Sept 24, 2008
NAME _
Exams for each section are given in the rooms listed. If you are in the wrong section, move to
the right one immediately. Circle your section number below.
SECTION
EXAMINATION
NUMBER
ROOM
1
DCC 308
2
DCC 3
PHYS-1200 PHYSICS II
EXAM 3-conflict
NAME_
Dec. 4, 2007
Please check the number next to your section below.
SECTION EXAMINATION
INSTRUCTOR
Section time
NUMBER ROOM
_1 DCC 308
SCOTT DWYER
MR 8:00 A.M.
_2 DCC 308
SCOTT DWYER
MR 10:00 A.M.
_3 West Hall
HEIDI