Open Query 3

#$&*

course Phy 121

Chapter 1 Problems 18-21 Your solution, attempt at solution. If you are unable to attempt a

solution, give a phrase-by-phrase interpretation of the problem along

with a statement of what you do or do not understand about it. This

response should be given, based on the work you did in completing the

assignment, before you look at the given solution.

003. `Query 3

*********************************************

Question: What do the coordinates of two points on a graph of position

vs. clock time tell you about the motion of the object? What can you

reason out once you have these coordinates?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The two coordinates on the graph it tells you how far the object moves

at a certain time. The slope of the graph represents the average

velocity for the interval of the 2 points. The rise is the change in

position and the run is the change in time.The position is on the

vertical axis and the time is on the horizontal axis.

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution: The coordinates a point on the graph include a position

and a clock time, which tells you where the object whose motion is

represented by the graph is at a given instant. If you have two points

on the graph, you know the position and clock time at two instants.

Given two points on a graph you can find the rise between the points and

the run.

On a graph of position vs. clock time, the position is on the 'vertical'

axis and the clock time on the 'horizontal' axis.

The rise between two points represents the change in the 'vertical'

coordinate, so in this case the rise represents the change in position.

The run between two points represents the change in the 'horizontal'

coordinate, so in this case the run represents the change in clock time.

The slope between two points of a graph is the 'rise' from one point to

the other, divided by the 'run' between the same two points.

The slope of a position vs. clock time graph therefore represents rise /

run = (change in position) / (change in clock time).

By the definition of average velocity as the average rate of change of

position with respect to clock time, we see that average velocity is

vAve = (change in position) / (change in clock time).

Thus the slope of the position vs. clock time graph represents the

average velocity for the interval between the two graph points.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique Rating: Ok

*********************************************

Question: 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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution: I would say we know that there is 8 significant figures

between 69 and 61. If will subtract them we would get 8. But the answer

would have one singificant figure.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

*********************************************

Question: 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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution: Possible units for position would be cm, meters,

millimeters. The standard unit is meters. The possible units for clock

time is seconds, minutes, years, microseconds. The standard unit is

seconds. The possible units for rate of change in position with respect

to clock time is meters/sec. millimeters/hour, kilometers/years, cm/sec

and many more.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

*********************************************

Question: `qQuery Principles of Physics and General College Physics:

Summarize your solution to Problem 1.19 (1.80 m + 142.5 cm + 5.34 * 10^5

`micro m to appropriate # of significant figures)

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: First you convert all the units to meters. Like 142.5cm

would be 1.425m and 5.34*10^5 micro m would be .534 meters. Then you

would take and add up all the numbers (1.80 m + 1.425 m +.534 m) and

your result would be 3.76 m.

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution: `a** 1.80 m has three significant figures (leading zeros

don't count, neither to trailing zeros unless there is a decimal point;

however zeros which are listed after the decimal point are significant;

that's the only way we have of distinguishing, say, 1.80 meter (read to

the nearest .01 m, i.e., nearest cm) and 1.000 meter (read to the

nearest millimeter).

Therefore no measurement smaller than .01 m can be distinguished.

142.5 cm is 1.425 m, good to within .00001 m.

5.34 * `micro m means 5.34 * 10^-6 m, so 5.34 * 10^5 micro m means (5.34

* 10^5) * 10^-6 meters = 5.34 + 10^-1 meter, or .534 meter, accurate to

within .001 m.

Then theses are added you get 3.759 m; however the 1.80 m is only good

to within .01 m so the result is 3.76 m. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

*********************************************

Question: For University Physics students: Summarize your solution to

Problem 1.31 (10th edition 1.34) (4 km on line then 3.1 km after 45 deg

turn by components, verify by scaled sketch).

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution: `a** THE FOLLOWING CORRECT SOLUTION WAS GIVEN BY A

STUDENT: The components of vectors A (2.6km in the y direction) and B

(4.0km in the x direction) are known.

We find the components of vector C(of length 3.1km) by using the sin and

cos functions.

Cx was 3.1 km * cos(45 deg) = 2.19. Adding the x component of the

second vector, 4.0, we get 6.19km.

Cy was 2.19 and i added the 2.6 km y displacement of the first vector to

get 4.79.

So Rx = 6.19 km and Ry = 4.79 km.

To get vector R, i used the pythagorean theorem to get the magnitude of

vector R, which was sqrt( (6.29 km)^2 + (4.79 km)^2 ) = 7.9 km.

The angle is theta = arctan(Ry / Rx) = arctan(4.79 / 6.19) = 37.7

degrees. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique Rating:

*********************************************

Question: A ball rolls from rest down a book, off that book and onto

another book, where it picks up additional speed before rolling off the

end of that book.

Suppose you know all the following information:

How far the ball rolled along each book.

The time interval the ball requires to roll from one end of each book to

the other.

How fast the ball is moving at each end of each book.

How would you use your information to determine the clock time at each

of the three points, if we assume the clock started when the ball was

released at the 'top' of the first book?

How would you use your information to sketch a graph of the ball's

position vs. clock time?

(This question is more challenging that the others): How would you use

your information to sketch a graph of the ball's speed vs. clock time,

and how would this graph differ from the graph of the position?

confidence rating #$&*: You would time the ball from the point of release

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

until the ending point and then divide the result by the length of the

books to get the average velocity. For the graph you would put speed on

the y axis and time on the x axis. I would use my measurments as my

points on my graph to show the area and slope of the graph. It would

differ because you are using different values on the graph.

"

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

*********************************************

Question: A ball rolls from rest down a book, off that book and onto

another book, where it picks up additional speed before rolling off the

end of that book.

Suppose you know all the following information:

How far the ball rolled along each book.

The time interval the ball requires to roll from one end of each book to

the other.

How fast the ball is moving at each end of each book.

How would you use your information to determine the clock time at each

of the three points, if we assume the clock started when the ball was

released at the 'top' of the first book?

How would you use your information to sketch a graph of the ball's

position vs. clock time?

(This question is more challenging that the others): How would you use

your information to sketch a graph of the ball's speed vs. clock time,

and how would this graph differ from the graph of the position?

confidence rating #$&*: You would time the ball from the point of release

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

until the ending point and then divide the result by the length of the

books to get the average velocity. For the graph you would put speed on

the y axis and time on the x axis. I would use my measurments as my

points on my graph to show the area and slope of the graph. It would

differ because you are using different values on the graph.

"

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

#*&!

&#Your work looks very good. Let me know if you have any questions. &#