Kaitlin Brantley
Global warming is basically an increase in the Earths atmosphere overall. This is supposedly caused by
something called a greenhouse effect and it is told that certain levels of carbon dioxide, and
chlorofluorocarbons are the cause of thi
Experiment 1: Newtons First Law
Procedure
1. Fill the container with about 4 inches of water.
2. Find an open space outside to walk around in with the container of water in your hands.
3. Perform the following activities:
a. Start with the water at rest (
Student Name: Kaitlin Brantley
Question Unit Conversion:
a. 46,756,790 mg = 46.75679 kg
b. 5.6 hours = 20,160 seconds
c. 13.5 cm = 5.314961 inches
d. 47 C =116.6F
1. For each of the following, convert each value into the designated units.
Question Signifi
Questions (use the below image to answer the questions):
1. During which period(s) is the person traveling in a positive direction?
0-2,12-15
2. During which period(s) is the person traveling in a negative direction?
4-7,9-10
3. During which period(s) is
Table 1: Applied force required to slide cup
Cup Material
Plastic
Styrofoam
Force Applied F1 m1 =
300 g water
Force Applied F2 m2 =
150 g water
Surface
Description
F2/ FN2
50
40
50
40
50
40
50
40
50
40
50
40
50
40
50
40
50
40
50
40
Avg: 50
Avg:40
Avg:50
A
Questions:
1. Draw a picture of a moving source and the waves surrounding it according to what you
observed in this experiment. How does the spacing of the wavefronts in front of the
source compare to those behind it?
I dont know how to draw a diagram on
Questions
1. Explain what caused the balloon to move in terms of Newtons Third Law.
The force of air affected both the straw and the balloon. The air rushing out, caused the
balloon to move forward , and the tape was used as a force as well between the ba
Student Name: Kaitlin Brantley
Experiment 1: Design an experiment
How many drops of water can fit on a penny? This experiment is to be designed by you!
Using any method you wish, conduct an experiment to answer this question. Next, see if
this number vari
Table 2: Coffee Filter Data
Procedure 1
1 Coffee Filter
2 Coffee Filters
Height of table (m)
.889
.889
Total Time (s) - Trail 1
1.35
.89
Total Time (s) - Trail 2
1.25
.79
Total Time (s) - Trail 3
1.33
.96
Total Time (s) - Trail 4
2.18
1.03
Total Time (s)
Exercise 2:
Determine which of the following observations are testable. For those that are testable:
Write a hypothesis and null hypothesis
What would be your experimental approach?
What are the dependent and independent variables?
What is your contro
30. Eq. 33-27 suggests that the slope in an intensity versus inverse-square-distance graph
(I plotted versus r 2 ) is P/4. We estimate the slope to be about 20 (in SI units) which
means the power is P = 4(30) 2.5 102 W.
29. (a) The upward force supplied by radiation pressure in this case (Eq. 33-32) must be
equal to the magnitude of the pull of gravity (mg). For a sphere, the projected area
(which is a factor in Eq. 33-32) is that of a circle A = r2 (not the entire surfa
28. The mass of the cylinder is m = ( D 2 / 4) H , where D is the diameter of the cylinder.
Since it is in equilibrium
Fnet = mg Fr =
HD 2 g D 2 2 I
= 0.
4
4 c
We solve for H:
H=
=
2I 2P 1
=
gc D 2 / 4 gc
2(4.60W)
[(2.6010 m) / 4](9.8m/s 2 )(3.0 108
25. (a) Since c = f , where is the wavelength and f is the frequency of the wave,
f=
c 2.998 108 m / s
=
= 10 108 Hz.
.
3.0 m
(b) The angular frequency is
= 2 f = 2 (1.0 108 Hz) = 6.3 108 rad / s.
(c) The angular wave number is
k=
2
2
=
= 2.1 rad / m.
3
25. (a) Since c = f , where is the wavelength and f is the frequency of the wave,
f=
c 2.998 108 m / s
=
= 10 108 Hz.
.
3.0 m
(b) The angular frequency is
= 2 f = 2 (1.0 108 Hz) = 6.3 108 rad / s.
(c) The angular wave number is
k=
2
2
=
= 2.1 rad / m.
3
26. We require Fgrav = Fr or G and solve for the area A: cGmM s (6.67 1011 N m2 / kg 2 )(1500 kg)(1.99 1030 kg)(2.998 108 m / s) A= = 2 . . 2 Id es 2(140 103 W / m2 )(150 1011 m) 2 = 9.5 105 m2 = 0.95 km2 . mM s 2 IA = , 2 d es c
25. (a) Since c = f , where is the wavelength and f is the frequency of the wave,
f=
c 2.998 108 m / s
=
= 10 108 Hz.
.
3.0 m
(b) The angular frequency is
= 2 f = 2 (1.0 108 Hz) = 6.3 108 rad / s.
(c) The angular wave number is
k=
2
2
=
= 2.1 rad / m.
3
24. (a) We note that the cross section area of the beam is d 2/4, where d is the diameter
of the spot (d = 2.00). The beam intensity is
I=
P
d 2 / 4
=
5.00 103 W
b gc
9
h
2.00 633 10 m
2
= 3.97 109 W / m2 .
/4
(b) The radiation pressure is
pr =
I 3.97 10
23. Let f be the fraction of the incident beam intensity that is reflected. The fraction
absorbed is 1 f. The reflected portion exerts a radiation pressure of
pr =
2 f I0
c
and the absorbed portion exerts a radiation pressure of
pa =
(1 f ) I 0
,
c
where
22. (a) The radiation pressure produces a force equal to
(1.4103 W/m 2 ) ( 6.37 106 m )
I
2
= 6.0 108 N.
Fr = pr ( R ) = ( Re ) =
8
2.998 10 m/s
c
2
2
e
(b) The gravitational pull of the Sun on Earth is
GM s M e ( 6.67 10
=
Fgrav =
2
d es
= 3.6 1022 N,
w
21. The plasma completely reflects all the energy incident on it, so the radiation pressure
is given by pr = 2I/c, where I is the intensity. The intensity is I = P/A, where P is the
power and A is the area intercepted by the radiation. Thus
2 (1.5 109 W )
18. From the equation immediately preceding Eq. 33-12, we see that the maximum value
of B/t is Bm . We can relate Bm to the intensity:
Bm =
2c 0 I
Em
=
,
c
c
and relate the intensity to the power P (and distance r) using Eq. 33-27. Finally, we
relate to w
19. Since the surface is perfectly absorbing, the radiation pressure is given by pr = I/c,
where I is the intensity. Since the bulb radiates uniformly in all directions, the intensity a
distance r from it is given by I = P/4r2, where P is the power of the
18. From the equation immediately preceding Eq. 33-12, we see that the maximum value
of B/t is Bm . We can relate Bm to the intensity:
Bm =
2c 0 I
Em
=
,
c
c
and relate the intensity to the power P (and distance r) using Eq. 33-27. Finally, we
relate to w
17. (a) The average rate of energy flow per unit area, or intensity, is related to the electric
2
field amplitude Em by I = Em / 2 0c , so
c
hc
hc
Em = 2 0cI = 2 4 107 H / m 2.998 108 m / s 10 106 W / m2
h
= 8.7 102 V / m.
(b) The amplitude of the magneti
16. (a) The expression Ey = Em sin(kx t) it fits the requirement at point P [it] is
decreasing with time if we imagine P is just to the right (x > 0) of the coordinate origin
(but at a value of x less than /2k = /4 which is where there would be a maximum,
2
15. (a) We use I = Em /20c to calculate Em:
c
hc
hc
h
Em = 2 0 I c = 2 4 107 T m / A 140 103 W / m2 2.998 108 m / s
.
= 103 103 V / m.
.
(b) The magnetic field amplitude is therefore
Bm =
Em 103 104 V / m
.
=
= 3.43 106 T.
8
c
2.998 10 m / s
14. (a) The power received is
Pr = (1.0 10
12
W)
(300 m )2 / 4
4 ( 6.37 10 m )
6
(b) The power of the source would be
c
hc
P = 4 r 2 I = 4 2.2 104 ly 9.46 1015 m / ly
2
= 1.4 1022 W.
LM 10 10 W OP
.
h M 4 .6.37 10 m P = 11 10
hQ
Nc
2
12
15
6
2
W.