Physics 121 Summer 2017 Solutions to Homework #01
page 1 of 29
This homework is worth 15 points.
1. The size of a hydrogen atom. The radius of a hydrogen atom (denoted a) depends on the mass
of an electron (M = 9.1 1031 kg), Plancks constant (
h = 1.055 1
Physics 121 Summer 2017 Solutions to Homework #04
page 1 of 24
PHYS 121: Solutions to Homework #04
June 19, 2017
Note: this homework is worth 15 points
Problem 1:
v0
block
R
block
R
bowl
top view
side view
A block of mass m is placed in motion on the insi
Physics 121 Summer 2016 Solutions to Homework #06
PHYS 121: Solutions to Homework #06
June 28, 2016
Note: this homework is worth 20 points as follows:
Problem 1
Problem 2
Problem 3
Problem 4
Problem 5
Problem 6
Approximate Lettergrade equivalents:
Score
Physics 121 Summer 2017 Document #13 Rotational Motion Review Sheet
page 1 of 8
PHYS 121: Rotational motion and Angular Momentum Review Sheet
June 21, 2017
Analogy Table:
We introduced the idea of rotational (spinning) motion, and how this motion could be
Physics 121 Summer 2017 Solutions to Homework #07
PHYS 121: Solutions to Homework #07
June 28, 2017
page 1 of 16
Physics 121 Summer 2017 Solutions to Homework #07
page 2 of 16
Problem 1: A Final Exam Problem From 2011:
0
A round marble with a given mass m
Physics 121 Summer 2017 Gravity and Systems Review Sheet
page 1 of 6
PHYS 121: Gravity, CenterofMass, and Systems Review Sheet
June 19, 2017
Force Menu:
For P121 at this point of the course we have a force menu. That is to say, when we go
about writing
Physics 121 Summer 2017
Torque and Angular Momentum
Figure 1: A particle of mass m moves in a circle. In what direction does the angular
momentum point?
Particle angular momentum. The angular momentum of a particle is defined as
L = mr v.
(1)
Here m is th
On the Nature of Knowledge
From Platos Theaetetus (201)
Socrates
Well, then, this at least calls for slight investigation; for you have a whole
profession which declares that true opinion is not knowledge.
Theaetetus
How so? What profession is it?
Socrate
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Lab: Circle (32 points)
Objectives
After you have completed this laboratory, you will be able to:
Accurately measure in centimeters using a ruler.
Graph data and write the equation of the line.
Explain what the slope and yintercept on your graph represen
Lab: Free Fall (26 points)
Objectives
After you have completed this laboratory, you will be able to:
Calculate the distance and displacement of a freely falling object, given the
time of flight.
Calculate the average velocity and instantaneous velocity of
Lab: Newtons First Law (26 points)
Objectives
After you have completed this laboratory, you will be able to:
Use Newtons first law.
Identify inertia in given situations.
Introduction
Isaac Newton is arguably the most influential scientist who contributed
Lab: Calorimeter (26 points)
Objectives
After you have completed this laboratory, you will be able to:
Describe the relationship between heat and Calories.
Calculate amount of Calories in food.
This lab involves using fire and an open flame. Have an adult
Lab: Conduction in Gases (25 points)
Objectives
After you have completed this laboratory, you will be able to:
Observe how heat is conducted through a gas.
Compare and contrast conduction and convection.
Introduction
Heat can be transferred through conduc
Tho lun LSCHTKT:
L thuyt bn tay v hnh
ca Adam Smith
Ni dung tho lun
Gii thiu v Adam Smith v l thuyt Bn tay v
hnh
L thuyt Bn tay v hnh trong thc tin
Cc u nhc im ca l thuyt Bn tay v
hnh
Gii php khc phc nhc im ti Vit Nam
I. Gii thiu v Adam Smith
v l thuy
9/05/2017
Example 2
The Tracey Corporation has a machining facility which specialises in jobs for
the aircraft components market. The company previously used a conventional
product costing system in which there were two direct cost categories
(Direct mate
FACULTY OF BUSINESS AND ECONOMICS
ACCG921
Managerial Accounting
Session 1 2017
Assessment Guide
Note this is NOT the Unit Guide
The official Unit Guide is available online and is accessed separately through iLearn
of this unit. The assessment guide should
L thuyt LI TH SO SNH RICARDO
1. Vi nt v tc gi:



D.Ricardo (17721823) sng trong thi k cch mng cng nghip, nghin cu nhiu lnh vc
Tc phm tiu biu: Nhng nguyn l c bn ca chnh sch kt v thu kha & Nh ng nguyn
l ca kinh t chnh tr hc
N/v c bn: xc lp cc quy lut p
Ly thuyt nn KT hn hp P.Samuelson
Ly thuyt nn kinh t hn hp cua P. Samuelson
1. Vai net v P. Samuelson
 P. Samuelson (19152009) la nha kinh t h oc ngi My, tac gia c ua
cun Kinh t h oc ni ting va la ngi ng u trng phai chinh hin
ai.
 Phng phap phn tich KT:
Ch ngha trng thng
HC THUYT TRNG THNG
1. Hon cnh ra i:
Kinh t x hi:
l t tng kinh t u tin ca giai cp t sn trong giai o n ph ng
thc sn xut phong kin tan r v ch ngha t bn ra i
Chuyn t nn kinh t hng ha gin n sang nn kih t th trng
L thi k tch lu nguyn thy ca c
V.1. Qu trnh xch li gn nhau ca cc t tng kinh t hin i
QA TRNH XCH LI CA CC T TNG KT HIN I
Cc hc thuyt ca ch ngha t do mi
I LT trng tin hin i M
II LT trng cung M
 Khi lg sx l kt qu ca chi ph, chi ph mang li kch thch kt
Nhim v ca nh nc l xy dng cc k cc yu
1:
1 l ch ngha trng thng cao vai tr nh nc v s dng phng php tru tng ha nghin
cu
2. trng phi trng nng ln u tin phn chia t bn thnh t bn bt bin v t bn kh bin
3 W.Petty v D.Ricardo cho rng sc lao ng l hng ha v ng h vic tr lng ti thiu
4 J. Keynes cho rng khng
APPENDIX 1
STUDENT EVALUATION OF MEMBER PARTICIPATION
In order to encourage equal participation on the part of all group members, each
group will complete and turn in an evaluation of the group member (only for the
Major Project group reports and presenta
Cu 1: (5) Tr li ng hay sai v gii thch ngn gn:
1. i tng nghin cu ca ch ngha trng thng l lnh vc sn xut.
2. A.Smith l ngi u tin nu quy lut lu thng tin t.
3. Theo K.Marx, khng hong kinh t ca t bn ch ngha l do s can thit qu su ca th trng.
4. Theo J.Keynes, cng
E = 0 Similarly (B n) D F = 0 (4) 16 (C n) E D = 0
or (A n) F E = 0 F + (Bn) D = 0 (5) E D + (C
n) = 0 Equation (5) has a nonzero solution only if E D C n F B n D A
n F E = 0 (6) This determinental equation is a cubic
in n and it is called characteristic
Using parallel axes theorem, I = IG + 9 M (h1 + h2 + h3) 2 (6) and
I = IG + 9 M (h2 + h3 2h1) 2 (7) I = 6 M (3 3 2 3 1 3 1 2 1 3 2 2
2 h1 + h + h + h h 3h h 3h h + 4h 2c + h h ) (7) IG = I 9 M
(h2 + h3 2h1) 2 (8) Put equation (8) in (6), I = I + 9 M (h1 +
between two principles. 4.17 PoincareCartan Integral Invariant : We
derive formula for W in the general case when the initial and terminal
instant of time, just like initial & terminal coordinates are not fixed but
are functions of a parameter . W() = 2
particles. M.I. of rod about GL = 3 Ma 2 M.I. of particles about GL =
ma2 + 0 + ma2 = 2ma2 As systems are equimomental, 2ma2 = 3 Ma 2
m = 6 M & M 2m = M 3 2M 3 M = So masses of particles at A, G,
B are 6 M, 3 2M, 6 M respectively. 32 A B C x x B D C a Ex
p (q q t ) & & qjdt W =
+ = 2t t 1 j j j 2 1 n j 1 j j q L dt d q L p q Ht &
qj dt (5) 75 where = = j 2 2 2 j 2 j 2 1 n j 1 j j
p q Ht p q H t j j q j p + H 1 t1 Now we know that H = L
j p j q&j H1 = L1 j j p j q& and H2 = L2 j 2 j 2 j p q&
In the spec
P(x,y,0) y m P(x,y,z) M.I. of this small element about GL = x y (x2
+ y2 ) M.I. of lamina = b b a a (x2 + y2 )dx dy = 4 b 0 a 0
2 (x + y2 ) dx dy = 4 = + + b 0 b o 2
3 a 0 2 3 ay dy 3 a xy dy 4 3 x = 4 b 0 3 3 3 ay y 3 a + =
3 4ab (a b b ) 3 4a 2 3 + = (a