Physical Science -- Sample
Exam #2
Spring 2002
Useful number:
Speed of light, c = 2.998x10^8 m/s
1. A 100 Watt light bulb glows brighter than a 25 Watt light bulb when attached to a 120 V ac source. Which bulb has the larger resistance?
(a) Calculate the resistance of the 25 Watt bulb.
P=VI so
I=P/V=25W/120V=0.208333 A. Then R=V/I=120V/0.208333A=576 Ohm
(b) Calculate the resistance of the 100 Watt
bulb.
P=VI so
I=P/V=100W/120V=0.833333 A. Then
R=V/I=120V/0.833333A=144 Ohm
2. If the two bulbs of problem 1 are both connected at the same time in series with the ac source,
(a) calculate the current through the source.
Total
R=720 Ohms. Then I=V/R=120V/720 Ohms =
0.1666667 A
(b)
Which bulb
glows brighter?
The 25W bulb is brighter
because both carry the same current and more energy is dissipated in the larger
resistance.
3. If the two bulbs of problem 1 are both connected at the same time in parallel with the ac source,
(a) calculate the current through the source.
Now the total resistance is given by
1/R=1/R1 + 1/R2, so 1/R=1/(576 Ohm ) + 1/(144 Ohm) = 0.0086806/Ohm and R=115.2
Ohm
Then I=V/R=120V/(115.2 Ohm) = 1.04A
Or
Both bulbs have the same voltage when
connected in parallel.
The current through the 25 W bulb is given by I=V/R = 120V/(576 Ohm) =
0.208333 A.
The current through the 100 W bulb is given by I=V/R = 120V/(144 Ohm) =
0.833333 A.
Then the total current through the source
is I = 0.208333 A + 0.833333 A = 1.04 A
(b) Which bulb glows brighter?
The 100W bulb is brighter because both have the same voltage across them, so the smaller resistor carries more of the current, and P=VI.
4. Explain in some detail a method by which we can create current in a wire without the use of a voltage source (including a battery).
Simply
move the wire across lines of a magnetic field. This is how a generator works.
5. A certain piano string carries waves at a speed of 308 m/s. In fundamental mode it produces a note of wavelength 1.4 meters. What is the frequency of the third harmonic?
The frequency of the third harmonic is three times the frequency of the
first harmonic (the fundamental).
Therefore f3 = 3 x speed/wavelength =
3 x (308 m/s) / (1.4 m) = 660 Hz
6. A 1056Hz tuning fork is sounded at the same time a piano note is played. You hear two beats per second. You tighten the piano string slightly, and repeat the process. Now you hear three beats per second. What is the piano note frequency?
The beat frequency is the difference between the two notes. The tuning fork has a frequency of 1056Hz,
so the piano could have a frequency of either 1054Hz or 1058Hz. When you tighten the piano string, it will
have a higher frequency, so since you now hear three beats, it must have been
higher than the tuning fork to begin with, so 1058Hz. (You could also answer 1059Hz,
which is the final frequency.)
7. A radio wave is bounced off of the surface of the Moon. It is detected back at the radio antenna 2.053 seconds later. What information does this tell us about the Moon?
This
tells us the distance to the moon.
Distance (there and back) = c x t = 2.998x10^8 m/s x 2.053s = 6.1549 x 10^8 m = 6.1549 x 10^5
km. So Distance to the Moon = 3.0774 x
10^5 km.
8. (a) Show and explain how lenses are able to collect light rays together to form an image.
Draw two rays from one point on the object. The ray parallel to the axis is bent so as to pass through the
far focus point. The ray through the
center is not bent at all. Where they
cross on the other side of the lens is where the image will appear.
(b) How can the image they produce be magnified?
Bring up a second lens and position it so that the primary image is
closer to the second lens than its focal length. The second lens then operates as a magnifying glass.
9. A piece of radium (half-life of 1620 years) is emitting an unsafe level of radiation. It is estimated that it will not be safe until it is only emitting 1/16 of the current radiation. How long will it be before the radium piece is safe to deal with?
After one half life, one half the radiation; after two, one fourth;
after three, one eighth; after four, one sixteenth. So 4 x 1620 years = 6,480
years.
10. (a) Briefly explain how stars create their energy.
Stars fuse 4 protons, the nuclei of hydrogen, into one alpha particle,
the nucleus of Helium-4.
(b) In the process 0.7% of the mass is lost. Using this information, how much energy would be created from 1kg of stellar material.
The energy produced is given by E = m c^2, where m is the
"missing" mass which has been converted to energy.
E = (0.007 x 1 kg) x (2.998 x 10^8 m/s)^2 = 6.29 x 10^14 Joules