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Unformatted text preview: Physical Science 303 Quiz #2, Sep. 25 2007 Name : 1. The area of a sphere is 4π R2 where R is the radius of that sphere. A tennis ball has a radius of 3.25cm and the radius of a basketball is four times as large. What is the
ratio of the surface area of a basketball to the surface area of a tennis ball? (8 pts)
16(=42) times larger because linear length scale is 4 times longer. 2. Convert the following numbers in the proper SI units. ( 8 pts ) time : 3 minutes and 15 seconds (2 pts) = 195s
time : 2 hours 10 seconds (2 pts)= 7210 s
distance : 1.3 mm (2 pts)= 0.0013 m
distance : 4.6 miles ( 1 mile = 1.609 km ) (2 pts) = 7401.4 m 3. A pendulum oscillated back and forth 15 times in 30 seconds. (12 pts) a. How long did it take for the pendulum to oscillate one complete cycle? (4 pts)
30 sec/15 times = 2 sec
b. Find the frequency of the pendulum? (4 pts)
f = 1/T = 1/ 2 sec = 0.5 1/sec(Hz) c. If the length of the pendulum will be shorten, what happen to the period and
the frequency? ( circle at the right comment ) (4 pts)
period : Shorter frequency : Smaller Longer
Larger No change
No change 4. Below graph shows the relationship between the period squared (T 2) and length (L) of a pendulum. Answer the following questions. (12 pts) a. Express the relationship between the period squared (T2) and the length (L) algebraically. (3pts)
T2 = a X L (a=0.2/5 s2/cm=0.04 s2/cm )
b. Express the relationship between the period (T) and the length (L) algebraically. (3pts).
T = √a X √L
c. Find the frequency of a 5cm long pendulum. (3pts) Read the period squared for 5cm.
T2 = 0.2 s2, T=√0.2 s= 0. 447 214 s
f=1/T= 2.24 Hz
d. Find the period of a 12cm pendulum. (3pts)
T = √ a X √ L = √ 0.04 X √ 12= 0.69 s ...
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 Spring '08
 MODI

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