Physics II
Resistivity of Water
Hailey Butler
1) Calculating A and A
Width
Depth (m) (m)
0.0181
0.1770
0.0156
0.0185
0.0182
0.0170
0.1770
0.1775
0.1770
0.1775
Depthavg
Widthavg
Areaavg
0.0175
0.1772
0.0031
D
W
A
0.000122
0.0005440
5 0.0000985
2) Inaccurat
RUNNING HEAD: EXPERIMENT 4: PROJECTILE MOTION
Experiment 4: Projectile Motion
PH2011-31
Marcus Koga
Javier Montes de Oca
Phillip Vue
Dr. Colton
Experiment Date: 1/5/16
Report Date: 1/12/16
EXPERIMENT 4: PROJECTILE MOTION
Koga 2
Abstract
The goal of this e
Experiment 5: Propagation of Uncertainties
PH2011-31
Marcus Koga, Javier Montes de Oca, Phillip Vue
Dr. Colton
Experiment Date: 01/12/16
Report Date: 01/19/16
Purpose:
The purpose of this experiment was to determine how uncertain measured quantities can i
Marcus Koga
Physics II
Dr. Masoud
5/4/16
Experiment 9- Mass of the Electron
1. Data Table of accelerating voltage, current causing electron to move up, current causing electron
to move down, average current, magnetic field, and magnetic field squared.
Va
Electron
9.1 x 10-31 kg
-1.6 x 10-19 C
Neutron
1.67 x 10-27 kg
Proton
1.67 x 10-27 kg
1.6 x 1019 C
Net force
q1 q2
F = k*( r 2 )
Fp =
Fp2
+
Fp5
F = q*E
Electric Field
kq
E = r2
E =E=
dv
dr
F
qo
V due to a point charge
kq
Vp = r
Acceleration
q
a= m E
Linea
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MILWAUKEE SCHOOL OF ENGINEERING
LABORATORY REPORT
7- Design Your Own Lab
PH- 2010
Section No. 11
Student Name
Lab Preformed: May 3, 2011
Due Date: May 10, 2011
Dr. Masoud
TEAM
Abstract:
The purpose of this experiment was to design your own lab when given
Ph 2010 041
April 6, 2011
MILWAUKEE SCHOOL OF ENGINEERING
Friction
Abstract: In this experiment we evaluated an object as it slid across a table top in order to
determine the effects of the changing of velocity on the coefficient of kinetic friction. The
MILWAUKEE SCHOOL OF ENGINEERING
LABORATORY REPORT
4- Friction
PH- 2010
Section No. 11
Student Name
Lab Preformed: March 29, 2011
Due Date: April 17, 2011
Dr. Masoud
TEAM
Abstract:
The purpose of this experiment was to determine the coefficient of kinetic
, Milwaukee School of Engineering
PH2010 Test#3 Spring 201 1
Name Fn 20x0 Segitc %\
Q1: A 10.0 kg crate is pulled up a rough incline with an initial speed of 1.3 m/s. The pulling
force is F=100.0 N parallel to the incline, which makes an angle of 150
Milwaukee School of Engineering
PH2010 Test#l Spring 201 1
Name: M
Ql: A car on a straight road starts from rest and accelerate at 2 m/s2 until it reaches a speed of
20 m/s. Then the car travels for 20 s at constant speed until the brakes are applied,
s
PH2010 Test#3 Eguation sheet SgrinoI 2011
szxzxl V=x2_xl v=lim£ vZ
tZtl AHOAt dt
_ vzv1 . Av dv
a: a=11m a:
tzt1 AHOAI dt
v 2 v0 +at v2 =1): +2a(xx0) x = x +v0t+éat2 V : v21)"
A A 6 A Asi116 A A 2 A 2 9 1 A-V
X cos y (x) +( y) tan 7
First Lab Report Complete Tables 1a, 1b, 2 & 3 in this workbook Sketch objects by hand after you print tables Answer questions below: 1. What are possible errors in this experiment? Please explain. 2. Why was it useful to measure all materials in bot
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 number
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
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
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. 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. (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