course Phy121
9/18/09 around 11:40 am
Questions related to q_a_vvvv
1. If we were to multiply the unit cm / sec by the unit sec / hr what unit would we get?
We would get cm/hr since when multiplies the seconds would cancel out.
2. If we were to divide the unit cm by the unit cm / sec what unit would we get?
We multiply by the reciprocal which would leave the units in sec since the cm would cancel out.
3. In this qa we began with the definition of average velocity. We have seen that this definition is consistent with our idea of speed as distance traveled divided by time required. However our definition of velocity can produce a negative result, which was not the case for speed. The negative result will, for example, occur when the change in position is negative while the change in clock time is positive.
• How is it that the change in position can be negative?
The change in position of an object can be negative but time however only flows forward.
4. Having defined average velocity, we used symbols to stand for the initial and final positions and clock times. What expression did this give us for the average velocity?
‘ds = s2-s1 ‘dt = t2-t1 vAve = ‘ds/’dt which would be (s2-s1) / (t2-t1)
5. Having defined average velocity, we then represented a pair of initial and final positions and clock times on a graph of position vs. clock time. What did we learn about the graph?
The slope of the graph is equal to the average velocity.
6. What are some possible units for position? What are some possible units for clock time? What therefore are some possible units for rate of change of position with respect to clock time?
Meters, centimeters, etc.
Seconds, minutes, hours, etc.
m/s, cm/s, m/hr, cm/hr, etc.
Questions related to text
1. What is 3.2 km + 340 meters, to the correct number of significant figures?
340 m would convert to .34km so we would then have 3.2km+.34km = 3.54 km
2. Find 1.80 m + 142.5 cm + 5.34 * 10^5 micro m, to the correct number of significant figures. You should know that 'micro' indicates multiplication by 10^-6, so a micro m is 10^-6 meter.
3.2255607 m
You have way too many significant figures. 1.80 m is known only to within .01 m, so your answer would be rounded to 3.23 m.
3. Find 3.84 kilograms - 3842 grams, to the appropriate number of significant figures.
3840g-3842g=-2g
Your answer could be expressed as .002 kg, and since 3.84 kg is significant only to .01 kg, the .002 kg is insignificant.
So to the appropriate number of significant digits, there is no difference; i.e., the difference is 0 kg.
Another way to think of this: 3.84 kg could mean anything between 3.835 kg and 3.845 kg. So the difference could be anything from -.007 kg to +.003 kg. You don't know whether the result is positive or negative.
Questions related to key systems
1. Pendulums of lengths 20 cm and 25 cm are counted for one minute. The counts are respectively 69 and 61. To how many significant figures do we know the difference between these counts?
The significant figures are to the number of significant figures in the problem so 2 significant figures.
This is so, but the difference in only 8 counts. Do we know 8 to two significant figures? The answer is that we do not.
This illustrates how the difference between two numbers can have fewer significant figures than the numbers themselves.
&#Your work looks good. See my notes. Let me know if you have any questions. &#