Why Science is Important Assignment
by Adriene Hill
Marketplace for Friday, August 23, 2013
STORY
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This smart grid can save power - and lives
Could it possibly be true that watch
Vector Worksheet
Much of the physical world can be described in terms of numbers. Examples of this are the
mass of an object, its temperature and its volume. These are called scalar quantities. But
some quantities also need a direction to fully describe i
Coulomb Law
Purpose
In this lab you will use the Coulomb Torsion Balance to show the inverse squared law for
electrostatic force between charges.
Equipment
Coulomb Balance and accessories, kilovolt power supply, cables (including probe and alligator
clip)
Measurement of Charge-to-Mass (e/m) Ratio for the Electron
Experiment objectives: measure the ratio of the electron charge-to-mass ratio e/m by
studying the electron trajectories in a uniform magnetic field.
History
J.J. Thomson first measured the charge-
Measurement of Electrical Resistance and Ohms Law
Objectives
In this experiment, measurements of the voltage across a wire coil and the current in the wire coil
will be used to accomplish the following objectives:
1. Definition of the concept of electrica
EQUATIONS
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Physics 220
Exam 1
3/17/2016
Name_
1. A car starts from rest and accelerates uniformly at 3.0 m/s2 toward the north. A second car
starts from rest 6.0 s later at the same point and accelerates uniformly at 5.0 m/s2 toward
the north.
a. How long after the
Physics 66 Lab Schedule
October 20th
October 27th
November 3rd
November 10th
November 17th
November 24th
December 1st
December 8th
Lab 12
Lab 13
Lab 10
Lab 17
Lab 19
Lab20
Lab 23rd
Conservation of Spring
Ballistic
Torque
Moment of Inertia
Pendulum
Mass on
Test #2 Extra Credit
1. A charge of 0.060 C moved vertically in the field of 0.090 T that is oriented 45
from the vertical. What speed must the charge have such that the force acting on
it is 10 N?
a. 3,500 m/s
b. 2,600 m/s
c. 1,500 m/s
d. 7,600 m/s
ANSWE
Extra Credit 1
1. A piece of plastic has a net charge of +6.00 C. How many more protons than electrons does this piece
of plastic have?
a. 6.781013
b. 4.501013
c. 1.601013
d. 3.751013
ANSWER: D
2. How much work is required to move an electron between two
Extra Credit #4
210
1. What is the resulting nucleus when
Pb undergoes alpha decay?
82
234
a.
U
92
206
b.
Hg
80
222
c.
Fr
87
210
d.
Pb
82
ANSWER: B
2. What is the resulting nucleus when
222
86
222
b.
73
222
c.
87
254
d.
90
a.
222
Rn undergoes beta-minus d
Extra Credit #3
1. A mirror has a focal length -15 cm. If an object is placed 10 cm in front of the mirror,
where will the image form?
a.) 7.5 cm behind the mirror
b.) 6 cm in front of the mirror
c.) 6 cm behind the mirror
d.) 7.5 cm in front of the mirro
Ch. 45
Page 1
CHAPTER 45 Astrophysics and Cosmology Note: A factor that appears in the analysis of energies is e2/4p0 = (1.60 1019 C)2/4p(8.85 1012 C2/N m2) = 2.30 1028 J m = 1.44 MeV fm. 1. Using the definition of the parsec, we find the equivalent dista
Ch. 44 Page 1
CHAPTER 44 Elementary Particles Note: A useful expression for the energy of a photon in terms of its wavelength is E = hf = hc/ = (6.63 1034 J s)(3 108 m/s)(109 nm/m)/(1.60 1019 J/eV); E = (1.24 103 eV nm)/ = (1.24 1012 MeV m)/.
1.
The total
Ch. 43 Page 1
CHAPTER 43 Nuclear Energy; Effects and Uses of Radiation Note: A factor that appears in the analysis of energies is e2/4p0 = (1.60 1019 C)2 /4p(8.85 1012 C2/N m2)= 2.30 1028 J m = 1.44 MeV fm.
1.
We find the product nucleus by balancing the
Ch. 42 Page 1
CHAPTER 42 Nuclear Physics and Radioactivity Note: A factor that appears in the analysis of energies is e2/4p0 = (1.60 1019 C)2/4p(8.85 1012 C2/N m2) = 2.30 1028 J m = 1.44 MeV fm.
1.
To find the rest mass of an particle, we subtract the res
Chapter 41
p. 1
CHAPTER 41 Molecules and Solids Note: At the atomic scale, it is most convenient to have energies in electron-volts and wavelengths in nanometers. A useful expression for the energy of a photon in terms of its wavelength is E = hf = hc/ =
Chapter 40 p. 1
CHAPTER 40 Quantum Mechanics of Atoms Note: At the atomic scale, it is most convenient to have energies in electron-volts and wavelengths in nanometers. A useful expression for the energy of a photon in terms of its wavelength is E = hf =
Chapter 39 p. 1
CHAPTER 39 Quantum Mechanics Note: At the atomic scale, it is most convenient to have energies in electron-volts and wavelengths in nanometers. A useful expression for the energy of a photon in terms of its wavelength is E = hf = hc/ = (6.
Chapter 38 p. 1
CHAPTER 38 Early Quantum Theory and Models of the Atom Note: At the atomic scale, it is most convenient to have energies in electron-volts and wavelengths in nanometers. A useful expression for the energy of a photon in terms of its wavele
Ch. 36 p. 1
CHAPTER 36 Diffraction and Polarization 1. We find the angle to the first minimum from sin 1min = m/a = (1)(680 109 m)/(0.0345 103 m) = 0.0197, so 1min = 1.13. Thus the angular width of the central diffraction peak is ?1 = 21min = 2(1.13) = 2.
Ch. 35 p. 1
CHAPTER 35 The Wave Nature of Light; Interference 1. We draw the wavelets and see that the incident wave fronts are parallel, with the angle of incidence 1 being the angle between the wave fronts and the surface. The reflecting wave fronts are
Ch. 34 p. 1
CHAPTER 34 Lenses and Optical Instruments 1. (a) From the ray diagram, the object distance is about 3& focal lengths, or 250 mm.
FI O F
(b) We find the object distance from (1/do) + (1/di) = 1/f; (1/do) + (1/88.0 mm) = 1/65.0 mm, which gives d
Ch. 33
p. 1
CHAPTER 33 Light: Reflection and Refraction 1. (a) The speed in ethyl alcohol is v = c/n = (3.00 108 m/s)/(1.36) = (b) The speed in lucite is v = c/n = (3.00 108 m/s)/(1.51) = 2.21 108 m/s. 1.99 108 m/s.
2.
We find the index of refraction from
Chapter 32 p. 1
CHAPTER 32 Maxwells Equations and Electromagnetic Waves 1. The electric field between the plates depends on the voltage: E = V/d, so dE/dt = (1/d) dV/dt = (1/1.3 103 m)(120 V/s) = 9.2 104 V/m s. The displacement current is ID = 0A (dE/dt)
Chapter 31 page 1
CHAPTER 31 AC Circuits 1. (a) The reactance of the capacitor is XC1 = 1/2pf1C = 1/2p(60 Hz)(7.2 106 F) = 3.7 102 . (b) For the new frequency we have XC2 = 1/2pf2C = 1/2p(1.0 106 Hz)(7.2 106 F) = 2.2 102 . We find the frequency from XL =
Chapter 30, p. 1
CHAPTER 30 Inductance; and Electromagnetic Oscillations 1. The magnetic field of the long solenoid is essentially zero outside the solenoid. Thus there will be the same linkage of flux with the second coil and the mutual inductance will b