Query03

#$&*

course Phy 241

550pm 9/5/2012

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

.

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

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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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Two seperate points on a graph of position vs clock time tell you the velocity of an object at two distinct times. We can reason out a number of things including acceleration, assuming constant acceleration. We can also reason out average velocity, intial velocity and final velocity. For instance two point (t_1, s_1) and (t_2, s_2) on a graph of position vs time. the first point has a velocity of s_1/t_1 which is the slope at that particular point. You can find the average velocity between the two point by using the equation v_ave = 'ds/'dt = (s_2-s_1)/ (t_2- t_1). In order to find acceleration multiply the average velocity by 2 and then divide by the 'ds. This tell you how fast the object is speeding up, assuming a constant acceleration.

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

Rise over run is the slope which is very much the same as 'ds/'dt. It is easy to see that the two are very similar.

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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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution:

The least amount of significant figures in the given problem is 2 sig figs. The difference between the two counts is (69 - 61) = 8. Normally you would leave 2 significant figures in the solution because the problem set has 2 sig figs

but the uncertainty was +-1, so therefore the answer must be 8, rather than 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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution:

units for position can be meters, centimeters, kilometers, inches, feet, miles, etc.

units for clock time can be seconds, milliseconds, minutes and hours, etc.

units for rate of change of position with respect to clock time can by m/s or cm/s or feet/minute etc.

Be sure not to mix SI with English units

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)

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:OK

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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

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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).

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

A_y = 2.6km

B_x = 4.0km

C _x = 3.1cos(45) + 4.0km = 5.6285 km

Find out more about calculator

C_y = 3.1sin(45) + 2.6km = 4.2738 km

R = sqrt(C_y^2 + C_x^2)

= sqrt(18.27 + 31.68)

= 7.067 km

theta = arctan (y/x)

= 37.2 degrees

confidence rating #$&*:OK

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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):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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your Solution:

The least amount of significant figures in the given problem is 2 sig figs. The difference between the two counts is (69 - 61) = 8. Normally you would leave 2 significant figures in the solution because the problem set has 2 sig figs

but the uncertainty was +-1, so therefore the answer must be 8, rather than 8.0

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:

units for position can be meters, centimeters, kilometers, inches, feet, miles, etc.

units for clock time can be seconds, milliseconds, minutes and hours, etc.

units for rate of change of position with respect to clock time can by m/s or cm/s or feet/minute etc.

Be sure not to mix SI with English units

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:

confidence rating #$&*:OK

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

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

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

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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).

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

A_y = 2.6km

B_x = 4.0km

C _x = 3.1cos(45) + 4.0km = 5.6285 km

Find out more about calculator

C_y = 3.1sin(45) + 2.6km = 4.2738 km

R = sqrt(C_y^2 + C_x^2)

= sqrt(18.27 + 31.68)

= 7.067 km

theta = arctan (y/x)

= 37.2 degrees

confidence rating #$&*:OK

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

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

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):OK

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

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:

The least amount of significant figures in the given problem is 2 sig figs. The difference between the two counts is (69 - 61) = 8. Normally you would leave 2 significant figures in the solution because the problem set has 2 sig figs

but the uncertainty was +-1, so therefore the answer must be 8, rather than 8.0

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:

units for position can be meters, centimeters, kilometers, inches, feet, miles, etc.

units for clock time can be seconds, milliseconds, minutes and hours, etc.

units for rate of change of position with respect to clock time can by m/s or cm/s or feet/minute etc.

Be sure not to mix SI with English units

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:

confidence rating #$&*:OK

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

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

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):OK

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

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:

A_y = 2.6km

B_x = 4.0km

C _x = 3.1cos(45) + 4.0km = 5.6285 km

Find out more about calculator

C_y = 3.1sin(45) + 2.6km = 4.2738 km

R = sqrt(C_y^2 + C_x^2)

= sqrt(18.27 + 31.68)

= 7.067 km

theta = arctan (y/x)

= 37.2 degrees

confidence rating #$&*:OK

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

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

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):OK

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Self-critique Rating:OK

#*&!#*&!

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