Query 3

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course Phy 121

6/15 at 5:00 PM

003. `Query 3

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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?

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Your solution: The coordinates of two points on a graph of position vs. clock time tells you where the object is at (position) at a certain time. Once you have these coordinates you can find the slope of the graph which is change in y / change in x. Therefore, we will find the vAve of the graph because the slope is the same as finding ‘ds / ‘dt.

confidence rating #$&*: 3

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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.

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Self-critique (if necessary): OK

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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?

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Your Solution:

The difference between these counts is 69 - 61 = 8. But it will be to two significant figures since both counts were two significant figures; therefore we should write it as 8.0.

confidence rating #$&*:

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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?

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Your Solution:

Position units: cm, meters, miles

Clock time units: seconds, minutes, hours

Rate of change units: cm/s, meters/min, miles/hr

confidence rating #$&*:

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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)

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Your solution:

Convert everything to meters first.

142.5 cm * (10^-2 m/ 1 cm) = 1.425 m

(5.34 * 10^5 ‘microm) * (10^-6 m / ‘microm) = .534 m

(1.80 m + 1.425 m + .534 m) = 3.76 m

confidence rating #$&*: 3

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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. **

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Self-critique (if necessary): I should have explained the reason why the answer was in 3 significant figures (because 1.80 m is).

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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).

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Your solution:

confidence rating #$&*:

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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. **

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Self-critique (if necessary): Not required to do this problem since I am in Phy 121.

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Self-critique Rating: OK

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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?

Keep in mind that average velocity = ‘ds / ‘dt. We know the total time interval it took to travel both books; and, by adding the 2 values of distance traveled per book, we know the total distance the ball rolled along both books. We use this to find the overall average speed (total distance travelled / time elapsed). By using this average speed we can determine how long it took the ball to travel down each individual book: average speed / distance for book 1 = time interval for book 1 and average speed / distance for book 2 = time interval for book 2. Both of these time intervals should equal the total time interval for both books when added together.

We would use this information to graph the ball’s position vs. clock time. Clock time would be on the x axis and position would be on the y axis. The slope of the line would = the change in position / change in clock time. Therefore this slope is the same thing as average velocity.

To graph the ball’s speed vs. clock time, you need to know average velocity per book. To find average velocity for book 1: (total distance ball rolled along both books - distance ball rolled along book 1) / time interval for book 1. To find average velocity for book 2: (total distance ball rolled along both books - distance ball rolled along book 2) / time interval for book 2. Clock time would be on the x axis and speed would be on the y axis. The slope of the line would = change in speed / change in clock time. Therefore, the slope is the same as finding the average rate of change of speed with respect to clock time.

confidence rating #$&*: 2

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Self-critique (if necessary):

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Self-critique rating:

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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?

Keep in mind that average velocity = ‘ds / ‘dt. We know the total time interval it took to travel both books; and, by adding the 2 values of distance traveled per book, we know the total distance the ball rolled along both books. We use this to find the overall average speed (total distance travelled / time elapsed). By using this average speed we can determine how long it took the ball to travel down each individual book: average speed / distance for book 1 = time interval for book 1 and average speed / distance for book 2 = time interval for book 2. Both of these time intervals should equal the total time interval for both books when added together.

We would use this information to graph the ball’s position vs. clock time. Clock time would be on the x axis and position would be on the y axis. The slope of the line would = the change in position / change in clock time. Therefore this slope is the same thing as average velocity.

To graph the ball’s speed vs. clock time, you need to know average velocity per book. To find average velocity for book 1: (total distance ball rolled along both books - distance ball rolled along book 1) / time interval for book 1. To find average velocity for book 2: (total distance ball rolled along both books - distance ball rolled along book 2) / time interval for book 2. Clock time would be on the x axis and speed would be on the y axis. The slope of the line would = change in speed / change in clock time. Therefore, the slope is the same as finding the average rate of change of speed with respect to clock time.

confidence rating #$&*: 2

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Self-critique (if necessary):

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Self-critique rating:

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