Query 3

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course PHY 121

6/13 4

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:

From two points on a graph, you can determine the units of length per unit of time for

each point, so if you had (x1, y1) and (x2, y2). The velocity of the object at the

first point is x1/y1 and for the second point it is x2/y2. You can take those points to

determine the average velocity (x1 + x2)/2 during that time. You can use that average to

find out how far the object traveled in that time (vAve * time = distance covered). You

can determine the acceleration rate by dividing the change in velocity by the time

interval. You can use that acceleration rate to find out how much distance the object

covered in each time interval.

confidence rating #$&*: 2

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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: 3

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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:

To two significant figures. 25, 69, and 61 all have 2 significant figures. If you

think of 20 as plus or minus one, then it has 2 signficant figures, too.

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:

Some possible ways to measure position are mm, cm, m, km, in, ft, yd, miles.

Some possible ways to measure clock time are seconds, minutes, hours.

You could therefore measure mm/s, cm/s, cm/m, km/hr, in/s, in/m, ft/s, ft/m, ft/hr,

yd/s, yd/m, yd/hr, or mi/hr.

confidence rating #$&*:

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Question: What fraction of the Earth's diameter is the greatest ocean depth?

What fraction of the Earth's diameter is the greatest mountain height (relative to sea

level)?

On a large globe 1 meter in diameter, how high would the mountain be, on the scale of

the globe? How might you construct a ridge of this height?

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Your solution:

Earth's diameter: 12,756 kilometers or 12,756,000 meters

Greatest ocean depth: 10,924 meters; fraction of diameter: 8/10,000

Greatest mountain height: 8850 meters; fraction of diameter: 7/10,000

The mountain would be .8 mm high. I imagine that you would make a ridge that is a little

less than a mm high.

confidence rating #$&*: 2

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Given Solution:

The greatest mountain height is a bit less than 10 000 meters. The diameter of the

Earth is a bit less than 13 000 kilometers.

Using the round figures 10 000 meters and 10 000 kilometers, we estimate that the ratio

is 10 000 meters / (10 000 kilometers). We could express 10 000 kilometers in meters,

or 10 000 meters in kilometers, to actually calculate the ratio. Or we can just see

that the ratio reduces to meters / kilometers. Since a kilometer is 1000 meters, the

ratio is 1 / 1000.

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Self-critique (if necessary):

First of all, I didn't round enough. I used exact numbers to start. When I used the

numbers given in this explanation, I was able to get the 1/1000 ratio.

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Self-critique Rating: 2

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

solution to the following:

Find the sum

1.80 m + 142.5 cm + 5.34 * 10^5 `micro m

to the appropriate number of significant figures.

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Your solution:

First I changed the cm to m by dividing by 100, so I got 1.425 m. Then I took the

10^5'micro m, and changed that to 10^5*10^-6, which would be 10^-1. That makes the 5.34

* 10^-1 or .534 m.

Next, I added the numbers: 1.8 m + 1.425 m + .534 m, which gave me 3.759 m.

I needed 3 signtificant figures for this answer, so it became 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): OK

I had some trouble with this when I did it in the book, but I was able to figure it out

and check it before I did it here.

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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):

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Self-critique Rating:

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Question:

A ball rolls from rest down a book, off that book and smoothly 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.

The acceleration on each book is uniform.

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

points (top of first book, top of second which is identical to the bottom of the first,

bottom of second book), 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?

It says that you know the time interval the ball requires to roll from one end of each

book to the other. The first point time is ""0"". The time to get from one end to the

other of book 1 is given. If the elapsed time given is for the 2nd book only, you can

add the two times together to get the total elapsed time. If the 2nd piece of

information is the total time, you can subtract the time for the first book from the

total to find the time for the 2nd book.

You know how far the ball traveled along each book, so you can take the amount of time

it took to travel that distance and get its velocity at that point. There would be

three data points plotted on the graph. The first would be at (0,0). This is because

the ball was at rest and the timer hadn't started yet. The second point would show how

far the ball had gone at the end of the first book. The third point would show how far

the ball had gone when it reached the end of the second book. You could also use the

graph to determine the acceleration rate of the ball as it traveled down the books.

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 smoothly 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.

The acceleration on each book is uniform.

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

points (top of first book, top of second which is identical to the bottom of the first,

bottom of second book), 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?

It says that you know the time interval the ball requires to roll from one end of each

book to the other. The first point time is ""0"". The time to get from one end to the

other of book 1 is given. If the elapsed time given is for the 2nd book only, you can

add the two times together to get the total elapsed time. If the 2nd piece of

information is the total time, you can subtract the time for the first book from the

total to find the time for the 2nd book.

You know how far the ball traveled along each book, so you can take the amount of time

it took to travel that distance and get its velocity at that point. There would be

three data points plotted on the graph. The first would be at (0,0). This is because

the ball was at rest and the timer hadn't started yet. The second point would show how

far the ball had gone at the end of the first book. The third point would show how far

the ball had gone when it reached the end of the second book. You could also use the

graph to determine the acceleration rate of the ball as it traveled down the books.

confidence rating #$&*: 2"

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Self-critique (if necessary):

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Self-critique rating:

#*&!

&#Good work. Let me know if you have questions. &#