Class presentation slides for Physical Science 50 (second quarter) Spring Semester 2009

1. Aristotle
Natural Motion

2. Galileo
Experimentation


3. Newton
Newton’s Laws of Motion, Gravity


4. Linear Motion
distance
SPEED = time


5. Linear Motion
displacement
VELOCITY = time


6. Earth -> Sun
1.497 x 1011m
distance
velocity = 3 x 108m/s
 Time =
= 498.9 seconds
= 8.3 minutes

7. Sun -> Mars
2.28 x 1011m
d
3 x 108m/s
t = v=
= 760.3 seconds
= 12.67 minutes

8. Acceleration
change in velocity
 Acceleration = time

9. Deceleration
= Negative Acceleration


10. Saleen S7 Turbo
vf– vi 60 mph – 0 mph mph
a= t = 3.3 s = 18.2 s
= 0.005 mi/s2

11. Gravitational Acceleration

12. Free Fall
Acceleration due to GRAVITY

Mass doesn’t matter


13. Gravitational Acceleration
 Velocity increases at a constant rate
vf - vi
a= t
20m/s - 10 m/s
= 1 second
= 9.8 m/s2
= 32 ft/s2

14. Distance Fallen
Distance = Velocityaverage x time

v i + vf
= 2 x t
0 + gt
= 2 x t
= 1/2 gt2

15. Transamerica Building
46 g golf ball dropped

from top of Transamerica
building
 7.35 seconds to hit
ground
d = ????


16. Transamerica Building
46 g golf ball dropped

from top of Transamerica
building
 7.35 seconds to hit
ground
 d = 1/2gt2
= (0.5)(9.8 m/s2)(7.35s)2
= 265 m
= 853 ft

17. g and elevation
 ATT vs. Coors Field
sea level 5280 ft
9.8185 m/s2 9.8136 m/s2

18. G force
Extreme Acceleration

> 1g
Breathing difficulties

blackout

19. Gravity

20. Newton’s Laws of Motion
Every body continues in a state of rest
1)
or in uniform motion in a straight line
unless compelled to change that state
by forces impressed upon it.
INERTIA

21. Inertia

22. Newton’s Laws of Motion
2) The acceleration produced by a Force
acting upon on body is directly
proportional to the mass of the body.
F=mxa

23. Force Units
F =m x a

= 2000 kg x 0.05 m/s2
= 100 kg m/s2 = 100 Newtons

24. Force of Gravity
F=mxa

Fg = m x g

Fg = WEIGHT


25. Mass vs. Weight
 60 kg person
Weight = m x g
Earth Moon Jupiter
(60 kg)(9.8m/s2) (60 kg)(1.67 m/s2) (60 kg)(25.9 m/s2)
588 N 100 N 1552 N

26. Weightlessness
“less” gravity

NOT
no gravity

27. Newton & Gravity
 1687
 Principia
“Every object in the

universe is attracted to
and attracts every other
object in the universe by
a force called gravity”

28. Gravitational Force
 Mutually
attractive force
 Inverse square
GM1M2
r2 where G
 Fg = = 6.67 x 10-11Nm2/kg2

29. g on Uranus?
Fg = m x g = GM1M2/r2

G Mplanet
g = rplanet2
(6.67 x 10-11 Nm2/kg2)(8.44 x 1025 kg)
g =(2.56 x 107 m)2
g = 8.59 m/s2

30. Size of Gravitational Force
(6.67 x 10-11 Nm2/kg2) (60 kg)(60 kg)
 Fg = (1 m)2
Fg = 2.4 x 10-7 N


31. Size of Gravitational Force
(6.67 x 10-11 Nm2/kg2) (6 x 1024 kg)(7.35 x 1022kg)
Fg =
 (3.83 x 108 m)2
Fg = 2 x 1020 N

32. Escape Velocity
d = 1/2gt2 for objects in free fall

2d = gt2 but t = v/g for free fall
2d/g = t2 = (v/g)2
2 dg = v2
√2dg = v = ESCAPE VELOCITY

33. Escape Velocity
vE = √2dg
= √2gREarth
= √2(9.8m/s2)(6.378 x 106 m)
= 11.18 X 103 m/s

34. How high will it go?
V2 = 2dg
V2/2g = d
(10 m/s)2
2 (9.8 m/s2) =d
d = 5.1 m

35. Newton’s Laws of Motion
In the absence of outside

forces, the total momentum of
a set of objects remains the
same no matter how the
objects interact with one
another.
p=mv


36. Conservation of Momentum
m vbefore = m vafter

(3000 kg)(10m/s) + (1000 kg)(0 m/s) = (3000 kg)v’ + (1000 kg) (15 m/s)
30,000 kg m/s = (3000 kg) v’ + 15,000 kg m/s
15,000 kg m/s = (3000 kg) v’ -------> v’ = 5 m/s

37. Conservation of Momentum
m vbefore = m vafter

(2000 kg) (8 m/s) + (50 kg) (0 m/s) = (2000 + 50 kg) v’
16,000 kg m/s ÷ 2050 kg = v’ = 7.8 m/s

38. Inelastic Collisions
(2 kg)(10 m/s) 3 kg
v’ =?
v = 0 m/s

39. Elastic Collisions
(2 kg)(10 m/s) 3 kg
v’ =?
v = -2 m/s

40. Angular Momentum
m1 x v1 x r1 = m2 x v2 x r2


41. Circular Motion
2 r = diameter

2πr = circumference

Magnitude of v doesn’t

change but direction
does

42. Centripetal Acceleration
v
a= t

vt - vb
= 1/2(2 r)/v
2v
= r/v
v2 mv2
ac = rFc = r

43. Applications
Spin cycle

washer

44. Car turning

45. The ROTOR

46. Centripetal Force
mball = 0.25 kg
r = 0.5 m
v = 2 m/s
Fc = ?

47. Centripetal Force
mball = 0.25 kg
r = 0.5 m
v = 2 m/s
(0.25 kg)(2 m/s)2
Fc = 0.5 m
= 2 Newtons

48. Roller Coaster Loop
Fc> Fg

mv2
r > mg
V2/r > g

49. Conservation of Energy
The Energy of the Universe

Is neither created nor destroyed;
It merely changes form.

50. Forms of Energy

51. Work and Power
W=Fxd

work
Power = time


52. Work
W=Fxd

W=mxgxd

W = (50 kg)(9.8m/s2)(5m)

= 2450 kg m2/s2
= 2450 Joules

53. Power
Work
 Power = time
2450 J
= 30 s
= 81.67 J/s
= 81.67 Watts

54. Kinetic Energy
Work = F x d = (m x a) x d
If d = 1/2at2 and v = at
then d = 1/2a(v/a)2 = 1/2v2/a
W = m a d = m a (1/2 av2/a) = 1/2mv2

55. Potential Energy
Potential Energy = m x g x h

56. Potential & Kinetic Energy
 PE + KE = Total Energy
 PEtop = KEbottom
mgh = 1/2mv2

2gh = v2

V = √2gh


57. Escape Velocity
vE = √2dg
= √2gREarth
= √2(9.8m/s2)(6.378 x 106 m)
= 11.18 X 103 m/s

58. Terminal Velocity
1 Kg book falls 1 meter
vf = ?
V = √2gh
= √2(9.8m/s2)(1m)
= √19.6 m2/s2 = 4.4 m/s

59. How tall is the incline?
V = 5.42 m/s

V = √2gh

Thus h = v2/2g
= (5.42 m/s)2/2(9.8 m/s2)
= 1.5 m

60. Roller Coaster

61. Pole Vault
World Record = 6 m

V = √2gh

= √2 (9.8m/s2)(6 m)
= 10.9 m/s

62. Heat transfer
Heat:

Energy
transferred between
2 systems in contact
and at different
temperatures

63. Temperature
Average kinetic energy

of atoms or molecules

64. Temperature scales

65. Biological Temperature Regulation
Human: 37 C = 98.6 F

Shivering, flushing


66. Biological Temperature Regulation
Birds

Animal fat layer


67. Specific Heat
The Amount of heat

energy necessary to
raise the
temperature
of a given
substance

68. Heat Energy
Heat Energy Q
Q=mC T
C in units of
Joules/g °C or
calories/g °C

69. Heat Capacity C

70. Calories
1 calorie:
amount of
heat needed
to raise temp
of 1.0 gram
H2O 1° C

71. Dietary Calories
1 Calorie = 1000 calories
100 Calories
100,000 calories!

72. Calorie burning

73. Heat Transfer
Wooden vs. Aluminum

Each pot contains 500 cal.
Each spoon 100 g
How hot will each one get?

74. Heat Transfer
 Aluminum spoon
Q = mC T = 500 cal
= (100g)(0.22cal/gºC) T
500 cal
T = (100g) (0.22cal/gºC)
= 22.73°C

75. Heat Transfer
 Wooden spoon
Q = mC T = 500 cal
= (100g)(0.58cal/gºC) T
500 cal
T = (100g) (0.58cal/gºC)
= 8.82°C

76. Identifying Unknowns
 A 400 g metal
w/3680 Joules of Heat
Temp 50°C -> 60°C
What is C?
Q
C=m T
= 3680 J/400g (10°C)
= 0.92 J/g °C

77. Limiting Heat Transfer
Cair = 0.17 cal/g°C


78. Refrigeration

79. Sound Energy
Result of Vibrations

Air molecules disturbed

Ear membrane vibration

Auditory nerve -> Brain


80. Hearing

81. Cochlea

82. Waves

83.  Wavelength: distance
between crests( )
 Frequency: # crests
pass a given pt/time (f)
 x f = velocity of sound
(m) (1/sec)
1/sec = HERTZ

84. Infrasound
f of elephant communication

if is 34 m?
fx = vs = 340 m/s
vs
Thus f = = 10 1/s = 10 Hz

85. Ultrasound
f of dolphin communication

if is 0.0034 m?
fx = vs = 340 m/s
vs
= 105 Hz
Thus f =

86. Audible Sound Waves
 Hearing range for
normal human:
20 - 20,000 Hz
 S,z,c sounds
vs.
 M,b sounds

87. Sound & Medium
Velocity of sound in different media
0º C air 331 m/s
Helium gas 970 m/s
Fresh water 1493 m/s
Salt water 1513 m/s
Granite 6000 m/s

88. Sound & Temperature
 vs = vo + (0.61 m/sºC)T

89. Hot Day vs. Cold Day
110° F vs. –30 ° F?

At 110° F = 43.33 °C
vs = 331 m/s + (0.61m/s °C) (43.3 °C)
= 357.43 m/s
At –30 ° F = –34.4 ° C
vs = 331 m/s + (0.61m/s °C) (–34.4 ° C)
= 310 m/s

90. Echolocation
Bat sends signal and

receives echo 0.7 s
later in 4° C cave --
how far is the wall?
vs = 331 m/s + (0.61m/s °C) (4°C) =333.4m/s
d = v x t = vs x t/2 = (333.4m/s)(0.35s) =116.7 m

91. Sound Intensity
Bel = 10 db

intensity doubling
(d2)2 I1
(d1)2 = I2
2
Intensity drops with (distance)

92. Sound Intensity

93. Intensity comparison
 Standing 2 m vs. 200 m
from jet taking off?
(200m)2 120 db
(2 m)2 x
40000/4 = 105 = 5 factors of 10
x = 70 db

94. Doppler Effect

95. Doppler Effect
 Approaching sound f
Vs
f’ = f VS - V
 Retreating sounds f
Vs
f’ = f VS + V

96. Doppler Example
Approaching at 30m/s:
340m/s
f’ = (440 Hz) 340 - 30 m/s
= 483 Hz
Retreating at 30 m/s:
340m/s
f’ = (440 Hz) 340 + 30 m/s
= 404 Hz

97. Medical Doppler
Blood flow velocity

F= 80000 Hz, f’ = 80020 Hz
f’ vs 1500 m/s
f = vs - v = 1.00025 = 1500 - v m/s
=> v = 0.375 m/s = 37.5 cm/s

98. Sonic Boom

99. Music
Pythagoras

 Octave = 2 f
 Harmonious frequencies = simpler ratio