1. (a) For a given value of the principal quantum number n, the orbital quantum number ranges from 0 to n 1. For n = 3, there are three possible values: 0, 1, and 2. (b) For a given value of , the magnetic quantum number m ranges from - to + . For =
Physics 222 Lab
Scalar Potential Fields and Vector Electric Fields
Equipment: 2 pages of field mapper grid (regular paper), low-voltage power supply, voltmeter, field
mapping kit including cork board, push pins, wires and sheets of conductive paper with e
Addendum to P222 L06.2 capacitance lab activity - This addendum describes how to make a new
column (in Logger Pro) representing the charge flowing from the capacitor of the capacitance lab
activity (L06.2). This new
Physics 222
RC Circuits Oscilloscope Lab
Equipment: function generator, oscilloscope, selected known resistors and capacitors, one 800 ml pyrex beaker,
one 600 ml pyrex beaker, salt, paper clip electrodes, banana cables, alligator clips, vernier caliper
O
Physics 222
Ohm's Law
Purpose: to investigate a common relationship between current and voltage
Equipment: Two resistors (each having resistance > 1000 ohms), 4 alligator clips, a variable voltage
supply, a voltmeter, an ammeter, 2 red cables and 2 black
Physics 222
Gauss Law Group Activity
Problems having spherical symmetry:
1) A positive point charge q1 is fixed at the center of a conducting spherical shell having inner
radius R1 and outer radius R2. The shell has a net positive charge Q2.
Find:
a) find
Physics 222 Group Activity
Electric Fields and Faraday Lines of Force
Drawing the Faraday lines of force for various charge distributions.
Rules:
1) Lines of force radially exit (arrows pointing away) positive point charges and radially enter (arrows
poin
Physics 222 Approved reference sheet for exam #2
!
! N kqi r
E = 2 E =
i=1 ri
V=
k ( dq ) r
r2
N
! ! qenc !
kqi
!
E i dA =
F = qE U = qV V =
"
0
i=1 ri
k ( dq )
!
V + V + k V
j
Physics 222 Approved reference sheet for exam #1
!
! N kqi r
E = 2 E =
i=1 ri
V=
k ( dq ) r
r2
N
! ! qenc !
kqi
!
E i dA =
F = qE U = qV V =
"
0
i=1 ri
k ( dq )
!
V + V + k V
j
Ch 13 - TYPES AND APPLICATIONS OF MATERIALS
13.1 (a) List the four classifications of steels. (b) For each, briefly describe the properties and
typical applications.
13.2 (a) Cite three reasons why ferrous alloys are used so extensively. (b) Cite three
ch
Homework Ch 8
8.23 (a) From the plot of yield strength versus (grain diameter)1/2 for a 70 Cu30 Zn cartridge
brass in Figure 8.15, determine values for the constants 0 and ky in Equation 8.7.
(b) Now predict the yield strength of this alloy when the avera
Homework Ch 10
10.8 Cite the phases that are present and the phase compositions for the following alloys:
(a) 90 wt% Zn-10 wt% Cu at 400C (750F)
(b) 75 wt% Sn-25 wt% Pb at 175C (345F)
(c) 55 wt% Ag-45 wt% Cu at 900C (1650F)
(d) 30 wt% Pb-70 wt% Mg at 425C
CH 15 COMPOSITES
Example 15.1
Example 15.2
15.8 A continuous and aligned fiber-reinforced composite is to be produced consisting of 30
vol% aramid fibers and 70 vol% polycarbonate matrix; the mechanical characteristics of these
two materials are as follow
Ch 11 - The Kinetics of Phase Transformations
Not stated in the homework problems: you may need to use Figures 11.18, 11.23, 11.31, 11.33, 11.36 and
Figure 7.30.
11.1 Name the two stages involved in the formation of particles of a new phase. Briefly descr
Ch 16 Corrosion
Ex. 16.1
16.4 (a) Compute the voltage at 25C of an electrochemical cell consisting of pure cadmium
immersed in a 2 10-3 M solution of Cd2+ ions and pure iron in a 0.4 M solution of Fe2+ ions.
(b) Write the spontaneous electrochemical react
Ch 12 Electrical Properties
12.2 A copper wire 100 m long must experience a voltage drop of less than 1.5 V when a current
of 2.5 A passes through it. Using the data in Table 12.1, compute the minimum diameter of the
wire.
12.11 At room temperature the el
Mai Abaidallah
Journal #2
9/3/2015
Politics of American Culture
The United States of America has a changing population and it still grows more diverse due to
the fact that not only a person can become an American citizen if he or she was born in the
Unite
Natural Sciences Division
Fullerton College
Physics 222 Spring 2017
Instructor: Brian Shotwell
Practice Problems, Chapter 28
April 17, 2017
1. What is the direction of the magnetic field at point P due to the two wires? Each wire is the same
distance d fr
1. Using the given conversion factors, we find (a) the distance d in rods to be d = 4.0 furlongs =
( 4.0 furlongs )( 201.168 m furlong )
5.0292 m rod
= 160 rods,
(b) and that distance in chains to be d =
( 4.0 furlongs )( 201.168 m furlong )
20.1
1. Charge flows until the potential difference across the capacitor is the same as the potential difference across the battery. The charge on the capacitor is then q = CV, and this is the same as the total charge that has passed through the battery.
1. (a) The charge that passes through any cross section is the product of the current and time. Since 4.0 min = (4.0 min)(60 s/min) = 240 s, q = it = (5.0 A)(240 s) = 1.2 103 C. (b) The number of electrons N is given by q = Ne, where e is the magnitu
1. (a) Eq. 28-3 leads to 6.50 10-17 N FB v= = = 4.00 105 m s . -19 -3 eB sin 160 10 C 2.60 10 T sin 23.0 .
c
hc
h
(b) The kinetic energy of the proton is
K=
2 1 2 1 mv = 167 10-27 kg 4.00 105 m s = 134 10-16 J. . . 2 2
c
hc
h
This is
1. (a) The magnitude of the magnetic field due to the current in the wire, at a point a distance r from the wire, is given by
B=
0i
2r
.
With r = 20 ft = 6.10 m, we have
c4 10 B=
hb 2 b6.10 mg
-7
T m A 100 A
g = 3.3 10
-6
T = 3.3 T.
(b
1. (a) The flux through the top is +(0.30 T)r2 where r = 0.020 m. The flux through the bottom is +0.70 mWb as given in the problem statement. Since the net flux must be zero then the flux through the sides must be negative and exactly cancel the tota
1. From the time dilation equation t = t0 (where t0 is the proper time interval,
= 1 / 1 - 2 , and = v/c), we obtain
= 1-
FG t IJ . H t K
2 0
The proper time interval is measured by a clock at rest relative to the muon. Specifically, t0 = 2.2
1. (a) Let E = 1240 eVnm/min = 0.6 eV to get = 2.1 103 nm = 2.1 m. (b) It is in the infrared region.
2. The energy of a photon is given by E = hf, where h is the Planck constant and f is the frequency. The wavelength is related to the frequency b
1. According to Eq. 39-4 En L 2. As a consequence, the new energy level E'n satisfies
En L = En L
FG IJ = FG L IJ H K H L K
-2
2
=
1 , 2
which gives L = 2 L. Thus, the ratio is L / L = 2 = 1.41.
2. (a) The ground-state energy is
( 6.63 10
1. The vector area A and the electric field E are shown on the diagram below. The angle between them is 180 35 = 145, so the electric flux through the area is
= E A = EA cos = (1800 N C ) 3.2 10-3 m cos145 = -1.5 10-2 N m 2 C.
2
(
)
2. We u