SETI : ASSIGNMENT 14
NAME: Valentina Ge
Score_/10
Open :
http:/certificate.ulo.ucl.ac.uk/modules/year_one/seti/drake_calculator.html
1) Put in the following values to the Drake EquationDrakes Original #s
R = 10
fp = 50
ne = 2.0
fl = 100
fi = 1
fc = 1
L =
Exam 2Waves/Particles
March 4, 2015
Modern Physics
Work quickly but carefully. For full credit, you must show all your work. If you cannot do part
of a question, assume a reasonable answer and continue on to the other parts.
Potentially useful information
Exam 1- Relativity
Modern Physics
Feb. 16, 2015
Work quickly but carefully. For full credit, you must show all your work. If you cannot do part
of a question, assume a reasonable answer and continue on to the other parts.
Potentially useful information:
!
Unit 1
1-D Kinematics
Distance vs. Displacement
o
o
o
o
Distance (s): the length of the trajectory
Displacement ( ): the overall change in position
Examples
In which cases is distance greater than / equal to /
smaller than displacement?
Average Speed vs.
Unit 3
Relative Motion & Circular Motion
Relative Motion
o Two identical eggs, one runs into the other. Which egg is
more likely to break?
Motion is relative.
Motion depends on reference frame.
Relative Motion
o General principle:
ra ,b ra ,c rc,b
va ,b v
Unit 5
Forces & Free-body Diagrams
Law of Universal Gravitation
o The gravitational force between two point masses lies
along the line joining the masses. Its magnitude is
proportional to the product of the masses and is inversely
proportional to the squa
Unit 7
WORK & KINETIC ENERGY
Work
o The work done by a force in moving an object over a
displacement is defined as
W F .d F .d .cos
F
d
o The SI unit of work is: J (joule)
1 J = (1 N)(1m) = 1 N.m
o Work can be positive, negative, or zero.
o This equation
Unit 6
FRICTION
Static Friction
o Static friction
o occurs at the contact surface of two objects when they
tend to slide against each other.
o has magnitude equal to the magnitude of the force
that is attempting to move the object, until it reaches a
maxi
Unit 4
Newtons Laws
Basic Concepts
o Force: the thing that is responsible for an objects change
in velocity.
o a vector quantity
o SI unit: newton (N)
o Inertia: the tendency of every object to maintain its state
of motion.
o Mass: the measure of inertia;
Unit 8
POTENTIAL ENERGY
Conservative Force & Potential Energy
o Conservative forces
o work is path-independent
o work on a closed path is zero
o Conservative force & potential energy:
xf
U Wcons F dx
xi
dU
Fx
dx
o Force is negative of the slope on U vs.
Valentina Ge
PHY 105 1
Reflective Paper 2
David Krieger
February 16, 2017
Scientific Method vs. Continental Drift
The continental drift theory was presented by the German meteorologist Alfred
Wegener. Throughout the process of exploring this hypothesis, w
Valentina Ge
PHY 105 1
David Kriegler
January 19, 2017
Scientific Method vs. Astrology
I have read the article about Astrology, so here comes my question: Is Astrology scientific?
There is claim of astrology that one's zodiac sign impacts one's ability to
Astronomy Ranking Task:
Phases of the Moon
Exercise #2
Description: The figure below shows a top view of the Sun, Earth and six different positions (A F) of the Moon during one orbit of Earth. Note that the distances shown for the Sun to Earth and for
Ear
28.48:(a) a =
F qvB sin ev 0 I = = m m m 2r (1.6 1017 C)(250,000 m s)(4 10 7 Tm A)(25 A) (9.11 10 31 kg)(2 )(0.020 m)
a=
= 1.1 1013 m s 2 , away from the wire. b)The electric force must balance the magnetic force. eE = eVB i E = vB = v 0 2r (250, 000 m/s)
19 4 I (1.60 10 C)(6.00 10 m s)(2.50 A) 28.47: F = qvB = qv 0 = 0 (0.045 m) 2r 2 19 = 1.07 10 N.
Let the current run left to right, the electron moves in the opposite direction, below the wire, then the magnetic field at the electron is into the page, and
F 0 I 2 1 1 0 I 2 = , upward. 28.20: On the top wire: = L 2 d 2d 4d On the middle wire, the magnetic fields cancel so the force is zero. F 0 I 2 1 1 0 I 2 = , downward. On the bottom wire: + = L 2 d 2d 4d
28.19:
B 1 , B
2
, B
3
B=
0 I ; r = 0.200 m for each wire 2r
B1 = 1.00 10 5 T, B2 = 0.80 10 5 T , B3 = 2.00 10 5 T Let be the positive z-direction. I 1 = 10.0 A, I 2 = 8.0 A, I 3 = 20.0 A B1z = 1.00 10 5 T, B2 z = 0.80 105 T, B3z = + 2.00 10 5 T B1z + B
28.17: The only place where the magnetic fields of the two wires are in opposite directions is between the wires, in the plane of the wires. Consider a point a distance x from the wire carrying I 2 = 75.0 A. Btot will be zero where B1 = B2 . 0 I1 I = 02 2
28.14: a) B0 =
0 I 2rB0 2 (0.040 m) (5.50 104 T) I= = = 110 A. 2r 0 0 I B 4 b) B = 0 , so B (r = 0.080 m) = 0 = 2.75 10 T, 2r 2 B0 B (r = 0.160 m) = = 1.375 10 4 T. 4
28.16: a)B = 0 since the fields are in opposite directions. I I I 1 1 b) B = Ba + Bb = 0 + 0 = 0 + 2ra 2rb 2 ra rb = (4 107 Tm A) (4.0 A) 1 1 0.3 m + 0.2 m 2
= 6.67 10 6 T = 6.67 T c)
B a ra
B b rb
Note that and B = Ba cos + Bb cos = 2 Ba cos tan = 5 = 1
28.18: (a)and (b)B = 0 since the magnetic fields due to currents at opposite corners of the square cancel. (c)
B = Ba cos 45 + Bb cos 45 + Bc cos 45 + Bd cos 45 I = 4 Ba cos 45 = 4 0 cos 45 2r r = (10 cm)2 + (10 cm)2 = 10 2 cm = 0.10 2 m B=4 (4 10 7 Tm A)
28.15: a) B =
0 I (800 A) =0 = 2.90 10 5 T, to the east. 2r 2 (5.50 m)
b)Since the magnitude of the earths magnetic filed is 5.00 10 5 T, to the north, the total magnetic field is now 30 o east of north with a magnitude of 5.78 10 5 T. This could be a pro
28.7: a) r = cos i + sin = cos(150)i + sin (150) = (0.866)i + (0.500) . j j j 3 b) dl r = ( dl i ) ( (0.866)i + (0.500) ) = dl (0.500)k = (5.00 10 m)k j I dl r I dl (0.500 m) (125 A)(0.010 m)(0.500 m) = 0 k= 0 k c) d B = 0 2 2 4 r 4 r 4 (1.20 m) 2 d B =
28.12: The total magnetic field is the vector sum of the constant magnetic field and the wires magnetic field. So: a)At (0, 0, 1 m): I (8.00 A) i = (1.0 10 7 T) i . B = B 0 0 i = (1.50 10 6 T) i 0 2r 2 (1.00 m) b)At (1 m, 0, 0): I (8.00 A) k B = B 0 + 0 k
28.8: The magnetic field at the given points is: I dl sin 0 (200 A) (0.000100 m) dBa = 0 = = 2.00 10 6 T. 2 2 4 r 4 (0.100 m) I dl sin 0 (200 A) (0.000100 m) sin 45 dBb = 0 = = 0.705 10 6 T. 2 2 4 r 4 2(0.100 m) I dl sin 0 (200 A) (0.000100 m) dBc = 0 = =
28.11: a)At the point exactly midway between the wires, the two magnetic fields are in opposite directions and cancel. b)At a distance a above the top wire, the magnetic fields are in the same 0 I 0 I 0 I 0 I 2 0 I k+ k= k+ k= k. direction and add up: B =