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course phy 121
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:
They tell you how far an object has traveled in a certain amount of time. Using the slope between these coordinated you can find the average velocity.
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
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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:
These values contain 2 sig figs. The difference would be 69-61 = 8, which is only one sig fig, so it could be written as 8.0 to show 2 sig figs.
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:
units for position are meters, cm, km, feet, mile, (10 meters, 10 cm, 10 km...ect) for time: seconds, millisecond, minute, hour. units for rate of change of position with respect to time would be cm/sec, feet/sec, miles/hr, ect
confidence rating #$&*:
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Question: What fraction of the Earth's diameter is the greatest ocean depth?
What fraction of the Earth's diameter is the greatest mountain height (relative to sea level)?
On a large globe 1 meter in diameter, how high would the mountain be, on the scale of the globe? How might you construct a ridge of this height?
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Your solution:
I am not sure where any of this material was mentioned/covered. (as far as given the data needed to solve)
confidence rating #$&*:0
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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): I am not sure where the data was found to solve these problems. Also, the given solution only explains one part of the above question.
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All the information used in solving this question is common knowledge, easily found.
Of course I wouldn't expect you to bring up all those dimensions on a test.
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Self-critique Rating:0
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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:
1.80 m + 1.425 m + .534 m = 3.759 = 3.76 m due to 1.80 m being to .01
confidence rating #$&*:2
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Given Solution:
`a** 1.80 m has three significant figures (leading zeros don't count, neither to trailing zeros unless there is a decimal point; however zeros which are listed after the decimal point are significant; that's the only way we have of distinguishing, say, 1.80 meter (read to the nearest .01 m, i.e., nearest cm) and 1.000 meter (read to the nearest millimeter).
Therefore no measurement smaller than .01 m can be distinguished.
142.5 cm is 1.425 m, good to within .00001 m.
5.34 * `micro m means 5.34 * 10^-6 m, so 5.34 * 10^5 micro m means (5.34 * 10^5) * 10^-6 meters = 5.34 + 10^-1 meter, or .534 meter, accurate to within .001 m.
Then theses are added you get 3.759 m; however the 1.80 m is only good to within .01 m so the result is 3.76 m. **
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Self-critique (if necessary):ok
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Question: Openstax: A generation is about one-third of a lifetime. Approximately howmany generations have passed since the year 0 AD?
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Your solution: if a lifetime is taken to be 70 years, 1/3 of 70 = 23.33
2013/23.33 = about 86 generations
confidence rating #$&*:3
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Given Solution:
A lifetime is about 70 years. 1/3 of that is about 23 years.
About 2000 years have passed since 0 AD, so there have been about
2000 years / (23 years / generation) = 85 generations
in that time
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Self-critique (if necessary): I realize it was an approximation, but I didnt not approximate all my values, just the answer.
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Self-critique Rating:3
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Question: Openstax: How many times longer than the mean life of an extremely unstable atomic nucleus is the lifetime of a human? (Hint: The lifetime of an unstable atomic nucleus is on the order of 10^(-22) s .)
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Your solution:
If the average life span is 70 years, that would be 70 years * 365 days per year * 24 hr/day * 60 min/hr * 60 sec/min = 2207520000 seconds = 2.2*10^9 / 10^-22 = 2.2*10^31
confidence rating #$&*:3
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Given Solution:
Assuming a 70-year human lifetime:
A years is 365 days * 24 hours / day * 60 minutes / hour * 60 seconds / minute = 3 000 000 seconds.
The number of seconds could be calculated to a greater number of significant figures, but this would be pointless since the 10^(-22) second is only an order-of-magnitude calculation, which could easily be off by a factor of 2 or 3.
Dividing 3 000 000 seconds by the 10^-22 second lifetime of the nucleus we get
1 human lifetime = 3 000 000 seconds / (10^-22 seconds / nuclear lifetime) = 3 * 10^28 nuclear lifetimes.
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Self-critique (if necessary):
I think in your solution, you forgot to multiply by 70 years. I believe you only found the total amount of seconds in 1 year.
@&
You are correct.
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Self-critique Rating:3
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Openstax: Calculate the approximate number of atoms in a bacterium. Assume that the average mass of an atom in the bacterium is ten times the mass of a hydrogen atom. (Hint: The mass of a hydrogen atom is on the order of 10−27 kg and the mass of a bacterium is on the order of 10−15 kg. )
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Question: For University Physics students: Summarize your solution to Problem 1.31 (10th edition 1.34) (4 km on line then 3.1 km after 45 deg turn by components, verify by scaled sketch).
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Your solution:
I dont understand what this is asking or how to solve it
confidence rating #$&*:0
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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):0
@&
This is a University Physics problem. You need not solve problems that weren't assigned for your course. You answer those questions which are not indicated for General College Physics or University Physics. If you are using the Giancoli text you don't need to do the problems marked "Openstax' which are for students using that alternative text . If you are using the alternative text, then you don't need to do the problems from Giancoli.
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Self-critique Rating:0
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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?
It says that we know how long it took for the ball to roll from one end of each book to the other. And since it was stated that the clock time started when the ball was first released from the first book, we can easily calculate at waht time the book was at each positon by knowing how long it took to roll down the book. If it stated at time zero on Book A, and took 2.5 seconds, than it was at the end of Book A/start of Book B at time 2.5 sec. The same process is used for book B
How would you use your information to sketch a graph of the ball's position vs. clock time?
Since we know how far it traveled, we would use that data to play the vertical access for positon, and the time intervals would be used for the horizontal axis.
(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?
Since we know how fast the ball was going at the end of each book, and we know that the acceleration is uniform, the average velocity of the ball must be 1/2 of the final velocity. However, the easiest way to plot this would be to make the vertical axis be velocity and the horizontal time. Plot a point at the origin since the b all is at rest at t0. The next point to plot would be the final velocity at the last clock time when the ball is at the end of the second book. We could also plot a point for the ball at the end of the first book, since we know fast it was traveling at the end of that book, and its time at that position. Since acceleration is uniform, this graph will be linear, so we can just connect a line through these points. The graphs differ because this graph is linear and the other was a curve. Their slopes would be very different. One slope would be constant (meaning it is changing at a constant rate) and the first graph, the slope would be changing at an changing rate(it would be increasing at an increasing rate)
confidence rating #$&*:3"
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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?
It says that we know how long it took for the ball to roll from one end of each book to the other. And since it was stated that the clock time started when the ball was first released from the first book, we can easily calculate at waht time the book was at each positon by knowing how long it took to roll down the book. If it stated at time zero on Book A, and took 2.5 seconds, than it was at the end of Book A/start of Book B at time 2.5 sec. The same process is used for book B
How would you use your information to sketch a graph of the ball's position vs. clock time?
Since we know how far it traveled, we would use that data to play the vertical access for positon, and the time intervals would be used for the horizontal axis.
(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?
Since we know how fast the ball was going at the end of each book, and we know that the acceleration is uniform, the average velocity of the ball must be 1/2 of the final velocity. However, the easiest way to plot this would be to make the vertical axis be velocity and the horizontal time. Plot a point at the origin since the b all is at rest at t0. The next point to plot would be the final velocity at the last clock time when the ball is at the end of the second book. We could also plot a point for the ball at the end of the first book, since we know fast it was traveling at the end of that book, and its time at that position. Since acceleration is uniform, this graph will be linear, so we can just connect a line through these points. The graphs differ because this graph is linear and the other was a curve. Their slopes would be very different. One slope would be constant (meaning it is changing at a constant rate) and the first graph, the slope would be changing at an changing rate(it would be increasing at an increasing rate)
confidence rating #$&*:3"
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Self-critique (if necessary):
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Self-critique rating:
#*&!
@&
Good.
Be sure to see my notes, especially concerning the problems you do and do not need to answer in the Queries.
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