Do NOT write on these sheets or take them with you! The next class needs them too!
PHYS1200 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:
PHYS1200 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 and Section:
PHYS1200 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
NAME: Section:
D («7% oxL
PHYS1200 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
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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
fBda=0
d
fEds= E¢~,:1
fBds= [101+ souoatCDE
47. We neglect air resistance for the duration of the motion (between "launching" and "landing"), so a = g = 9.8 m/s2 (we take downward to be the y direction). We use the equations in Table 21 (with y replacing x) because this is a = constant motion. (a)
48. 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 21 (with y replacing x) because this is constant acceleration motion. The ground level is
14. We denote the pulsar rotation rate f (for frequency). f = 1 rotation 1.55780644887275 103 s
(a) Multiplying f by the timeinterval t = 7.00 days (which is equivalent to 604800 s, if we ignore significant figure considerations for a moment), we obtain
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 21 (with y replacing x) because this is constant acceleration motion. We are placing the
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.
15. The time on any of these clocks is a straightline function of that on another, with slopes 1 and yintercepts 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
5. Various geometric formulas are given in Appendix E. (a) Expressing the radius of the Earth as
R = ( 6.37 106 m )(103 km m ) = 6.37 103 km,
its circumference is s = 2 R = 2 (6.37 103 km) = 4.00 104 km. (b) The surface area of Earth is A = 4 R 2 = 4 ( 6
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 24h 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 24h 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. 211 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. 215: y = (19.6 m/s)(2.0 s) +
1 2
(9.8 m/s2)(2.0 s)2 20 m .
We n
46. We neglect air resistance, which justifies setting a = g = 9.8 m/s2 (taking down as the y direction) for the duration of the fall. This is constant acceleration motion, which justifies the use of Table 21 (with y replacing x). (a) Noting that y = y y
13. (a) Presuming that a French decimal day is equivalent to a regular day, then the ratio of weeks is simply 10/7 or (to 3 significant figures) 1.43. (b) In a regular day, there are 86400 seconds, but in the French system described in the problem, there
39. The displacement (x) for each train is the "area" in the graph (since the displacement is the integral of the velocity). Each area is triangular, and the area of a triangle is 1/2( base) (height). Thus, the (absolute value of the) displacement for one
6. From Figure 1.6, we see that 212 S is equivalent to 258 W and 212 32 = 180 S is equivalent to 216 60 = 156 Z. The information allows us to convert S to W or Z. (a) In units of W, we have
50.0 S = ( 50.0 S) 258 W = 60.8 W 212 S
(b) In units of Z, we hav
7. The volume of ice is given by the product of the semicircular surface area and the thickness. The area of the semicircle is A = r2/2, where r is the radius. Therefore, the volume is V = r2 z 2 where z is the ice thickness. Since there are 103 m in 1 km
40. Let d be the 220 m distance between the cars at t = 0, and v1 be the 20 km/h = 50/9 m/s speed (corresponding to a passing point of x1 = 44.5 m) and v2 be the 40 km/h =100/9 m/s speed (corresponding to passing point of x2 = 76.6 m) of the red car. We h
8. We make use of Table 16. (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 8.33 102 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 21 is permitted. We start with Eq. 217 and denote the train's initial velocity as vt and the locomotive's velocity as v
45. We neglect air resistance, which justifies setting a = g = 9.8 m/s2 (taking down as the y direction) for the duration of the fall. This is constant acceleration motion, which justifies the use of Table 21 (with y replacing x).
2 (a) Starting the cloc
Physics 1200  Physics 11 NAME H 1 " " 3' "J
October 1, 2014 m
EXAM One SECTION NUMBER 9
Circle your section number below and write it above!
SECTION INSTRUCTOR TIME that the section Examination room
NUMBER meets
Emma
ME
=

n
im
PHYS1200 PHYSICS II NAME Z i MSW, a? 3
co Awe gamma
EXAMl March 4, 2015 Section _.Q_ KEY
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
Physics 1200
Lecture 09
Creating magnetic fields
Oersteds Demonstration
BiotSavart
Amperes Law
Physics 2  Persans
1
A moving charge
creates a magnetic field!
1. Field strength is proportional to
the velocity and charge
magnitude.
2. If v changes directi
Physics 2
Lecture 2 notes
Spring 2016
The Electric Field
1
The Idea of the Electric
Field
Coulombs Law tells
us the force between
two particles.
A more useful way of
thinking about this is
to consider that
charge 1 creates a
field to which charge
2 resp