#$&*
course phys201
1.If the position of an object changes from 34 cm at clock time 4.6 seconds to 87 cm at clock time 5.3 seconds, then during this interval what is the average rate of change of its position with respect to clock time?87-34/5.3-4.6 = 53/.7 75.7 avg velocity
2. If the velocity of an object changes from 12 cm/second at clock time 6.9 seconds to 20 cm/s at clock time 15.3 seconds, then what is the rate of change of its velocity with respect to clock time?
20/8.4 = 1.31 avg accel
3. What is your best estimate of the average velocity of the object in #2, for the given time interval?
8cm/8.4 =.95 avg velocity
4. If the average rate of change of position with respect to clock time during a certain interval is 24 meters / second, and if the interval lasts for 5 seconds, then what quantity can you determine by applying the definition of an average rate of change? Find this quantity and explain in detail how you found it.
24/5 = 4.8 beciase avg roc of position = 24 and clocktime = 5 so 24/5 = 4.8
5. What is wrong with saying the average velocity = position / clock time?
Because avg velocity is avg roc of position wrt cloctktime
Not a given position/ a given time
Text-related questions:
1. What is the percent uncertainty in a measured time interval of 3.4 seconds, given that the timing mechanism has an uncertainty of +- .1 second? What is the percent uncertainty in a time interval of .87 seconds, measured using the same mechanism? When using this mechanism, how does the percent uncertainty in measuring a time interval depend on the duration of that interval?
34% and 8.7% the percent of uncertainty vorrelates with theduration of the interval and significant figures
2. What is the uncertainty in the following reported measurements, and what is the percent uncertainty in each?
• 5.8 centimeters- 1.72
• 2350 kilometers - 4.2 x 10^-3
• 350. seconds 2.8 x 10^ -2
• 3.14 -3.18
• 3.1416 -3.18
2. What is the uncertainty in the area of a rectangle, based on reported length 23.7 cm and width 18.34 cm?
23.7x 18.34 = 434.658,
.1/434.658 x 100=
.023
3. (Principles of Physics students are invited to solve this problem, but are not required to do so): What is the approximate uncertainty in the area of a circle, based on a reported radius of 2.8 * 10^4 cm?
.1/2.8x10^4= 3.5x 10^-6=
3.5x10^-4
5. What is your height in meters, and your ideal mass in kilograms? How much uncertainty do you think there is in each, and why?
81 in= .12
205.74 cm=.048
2.0574 m = 4.86
Mass:
139 ibs- .07
63,049 g-1.5x10^-4
63.049 kg- .15
The smaller the unit the smaller percentage there is for erroe in measurements leaving smaller uncertainty, the opposite goes for the larger the unit of measurement.
"
&#This looks very good. Let me know if you have any questions. &#