Photo Mechanic 6 Manual Pdf

Photo Mechanic 6 Manual Pdf

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Photo Mechanic 6 Manual Pdf
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Размер:СкачатьSOLUTIONS MANUAL for elementary mechanics & thermodynamics
Professor John W. Norbury
Physics Department
University of Wisconsin-Milwaukee
P.O. Box 413
Milwaukee, WI 53201
Jutoh 1 30 ubkg download free. November 20, 2000
2
Contents
1
MOTION ALONG A STRAIGHT LINE
5
2
VECTORS
15
3
MOTION IN 2 & 3 DIMENSIONS
19
4
FORCE & MOTION - I
35
5
FORCE & MOTION - II
37
6
KINETIC ENERGY & WORK
51
7
POTENTIAL ENERGY & CONSERVATION OF ENERGY 53
8
SYSTEMS OF PARTICLES
57
9
COLLISIONS
61
10 ROTATION
65
11 ROLLING, TORQUE & ANGULAR MOMENTUM
75
12
OSCILLATIONS
77
13
WAVES - I
85
14
WAVES - II
87
15 TEMPERATURE, HEAT & 1ST LAW OF THERMODY-
NAMICS
93
16 KINETIC THEORY OF GASES
99
3
5
6
CHAPTER 1. MOTION ALONG A STRAIGHT LINE
1.The following functions give the position as a function of time:
i)x = A
ii)x = Bt
iii)x = Ct2
iv)x = D cos !t
v)x = E sin !t
where A; B; C; D; E; ! are constants.
A)What are the units for A; B; C; D; E; !?
B)Write down the velocity and acceleration equations as a function of time. Indicate for what functions the acceleration is constant.
C)Sketch graphs of x; v; a as a function of time.
SOLUTION
A)X is always in m.
Thus we must have A in m; B in m sec°1, C in m sec°2.
!t is always an angle, µ is radius and cos µ and sin µ have no units. Thus ! must be sec°1 or radians sec°1.
D and E must be m.
B)v = dxdt and a = dvdt . Thus
i) v = 0
ii) v = B
iii) v = Ct
iv) v = °!D sin !t
v) v = !E cos !t
and notice that the units we worked out in part A) are all consistent with v having units of m¢ sec°1. Similarly
i) a = 0
ii) a = 0
iii) a = C
iv) a = °!2D cos !t
v) a = °!2E sin !t
7
C)
i)
ii)
iii)
x
x
x
t
t
t
v
v
v
t
t
t
a
a
a
t
t
t
8
CHAPTER 1. MOTION ALONG A STRAIGHT LINE
iv)
v)
Photo Mechanic 6 Manual Pdf1
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x
0
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0
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a
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t
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2.The Øgures below show position-time graphs. Sketch the corresponding velocity-time and acceleration-time graphs.
x
x
x
t
t
t

SOLUTION
The velocity-time and acceleration-time graphs are:
v
t
t
a
10
CHAPTER 1. MOTION ALONG A STRAIGHT LINE
3.If you drop an object from a height H above the ground, work out a formula for the speed with which the object hits the ground.
SOLUTION
v2 = v02 + 2a(y ° y0)
In the vertical direction we have:
v0 = 0,
a = °g,
y0 = H,
y = 0.
Thus
v2 = 0 ° 2g(0 ° H)
=
2gH
) v
=
p
2gH
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1	MOTION ALONG A STRAIGHT LINE	5
2	VECTORS	15
3	MOTION IN 2 & 3 DIMENSIONS	19
4	FORCE & MOTION - I	35
5	FORCE & MOTION - II	37
6	KINETIC ENERGY & WORK	51
7	POTENTIAL ENERGY & CONSERVATION OF ENERGY 53
8	SYSTEMS OF PARTICLES	57
9	COLLISIONS	61
10 ROTATION	65
11 ROLLING, TORQUE & ANGULAR MOMENTUM	75
12	OSCILLATIONS	77
13	WAVES - I	85
14	WAVES - II	87
15 TEMPERATURE, HEAT & 1ST LAW OF THERMODY-
NAMICS	93
16 KINETIC THEORY OF GASES	99
3

i) v = 0	ii) v = B	iii) v = Ct
iv) v = °!D sin !t	v) v = !E cos !t

i) a = 0	ii) a = 0	iii) a = C
iv) a = °!2D cos !t	v) a = °!2E sin !t

1
0.5
v	0
-0.5 Minecraft 1 14 4.
-1	1	2	3	4	5	6 Buildbox 1 3 4 – drag and drop game builder.
0
t

SOLUTION
v2 = v02 + 2a(y ° y0)
In the vertical direction we have:
v0 = 0,	a = °g,	y0 = H,	y = 0.
Thus
v2 = 0 ° 2g(0 ° H)
=	2gH
) v	=	p
2gH