Victoria Clay
104053575
The following report will do an in depth assessment of the credibility, casual reasoning,
inference, and a study of the article Eating chicken Wings Makes Kids Aggressive.
Credibility
The articles credibility can be considered good
Victoria Clay 104053575
Assignment 3 Passages
(Instructions for this assignment are included in the document, Instructions for Assignment 3.
The following passages are those you will need to work with in order to complete the
assignment. There are 8 quest
The Impression of the article Commons erupts as opposition accuses Trudeau of
elbowing NDP MP is that the prime minister got very physical with a female opposition that
resulted in a serious uproar and it occurred before an important vote. This impression
Victoria Clay 104053575
Beth Pieterson is a highly credible source in regards to the claim that Wi-Fi signals
radiation has no health effects related to exposure. Beth has had ample opportunity to check
and research all of the data and findings being done
UNIVERSITY OF WINDSOR DEPARTMENT OF PHYSICS 64-140 Mid-term Test No. 1 Time: 50 minutes October 26, 2005
Name:
Student ID No:
DO NOT OPEN UNTIL INSTRUCTED.
NO CELL PHONES OR OTHER ELECTRONIC DEVICES ALLOWED. INSTRUCTIONS: Each student must write in the se
UNIVERSITY OF WINDSOR DEPARTMENT OF PHYSICS 64-140 Final Examination Time: 3 hours December 16, 2005
Name:
Student ID No:
DO NOT OPEN UNTIL INSTRUCTED.
NO CELL PHONES OR OTHER ELECTRONIC DEVICES ALLOWED. INSTRUCTIONS: Each student must write in the seat a
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 )
20.117 m chai
1. The x and the y components of a vector a lying on the xy plane are given by
ax = a cos , a y = a sin
where a =| a | is the magnitude and is the angle between a and the positive x axis. (a) The x component of a is given by ax = 7.3 cos 250 = 2.5 m. (b)
1. With speed v = 11200 m/s, we find
K= 1 2 1 mv = (2.9 105 ) (11200) 2 = 18 1013 J. . 2 2
2. (a) The change in kinetic energy for the meteorite would be
1 1 K = K f - K i = - K i = - mi vi2 = - 4 106 kg 15 103 m/s 2 2
(
)(
)
2
= -5 1014 J ,
or | K |= 5 1
1. (a) The center of mass is given by xcom = [0 + 0 + 0 + (m)(2.00) + (m)(2.00) + (m)(2.00)]/6.00m = 1.00 m. (b) Similarly, ycom = [0 + (m)(2.00) + (m)(4.00) + (m)(4.00) + (m)(2.00) + 0]/6m = 2.00 m. (c) Using Eq. 12-14 and noting that the gravitational e
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 total of
1. Conservation of momentum requires that the gamma ray particles move in opposite directions with momenta of the same magnitude. Since the magnitude p of the momentum of a gamma ray particle is related to its energy by p = E/c, the particles have the sam
1. If R is the fission rate, then the power output is P = RQ, where Q is the energy released in each fission event. Hence, R = P/Q = (1.0 W)/(200 106 eV)(1.60 10 19 J/eV) = 3.1 1010 fissions/s.
2. We note that the sum of superscripts (mass numbers A) must
1. Our calculation is similar to that shown in Sample Problem 42-1. We set K = 5.30 MeV=U = (1/ 4 0 )( q qCu / rmin ) and solve for the closest separation, rmin:
rmin
-19 9 q qCu kq qCu ( 2e )( 29 ) (1.60 10 C )( 8.99 10 V m/C ) = = = 4 0 K 4 0 K 5.30 106
1. The number of atoms per unit volume is given by n = d / M , where d is the mass density of copper and M is the mass of a single copper atom. Since each atom contributes one conduction electron, n is also the number of conduction electrons per unit volu
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 = 1 , t
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 J s ) h2 E1 = n2
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 by f = c,
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 = 2.2000 s. We a