Query 5

course Phy 201

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005. `query 5

Physics I

06-25-2008

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00:14:20

Intro Prob 6 Intro Prob 6 How do you find final velocity and displacement given initial velocity, acceleration and time interval?

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RESPONSE -->

To find the final velocity and displacement we first have to solve for the final velocity in the equation,

vf= v0 + a * `dt

from there we can find the average velocity by adding the final and initial velocity and dividing by 2. With the average velocity we can multiply that by time to obtain the displacement;

`ds = vAve * `dt

confidence assessment: 2

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00:15:07

** To find final velocity from the given quantities initial velocity, acceleration and `dt:

Multiply `dt by accel to get `dv.

Then add change in velocity `dv to init vel , and you have the final velocity**

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RESPONSE -->

ok that is a simpler way to do it!

self critique assessment: 2

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00:17:40

Describe the flow diagram we obtain for the situation in which we know v0, vf and `dt.

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RESPONSE -->

v0 and vf are on the first row and from each of the velocities lines will be connected to form vAve in the second row. `dt is also on the top row and is now connected to vAve and `ds in the second row.

confidence assessment: 2

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00:18:11

** The flow diagram shows us the flow of information, what we get from what, usually by combining two quantites at a time. How we get each quantity may also be included.

From vf and v0 we get `dv, shown by lines from vf and v0 at the top level to `dv. From vf and v0 we also get and vAve, shown by similar lines running from v0 and vf to vAve.

Then from vAve and `dt we get `ds, with the accompanying lines indicating from vAve and `dt to `ds, while from `dv and `dt we get acceleration, indicated similarly. **

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RESPONSE -->

ok

self critique assessment: 3

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00:20:11

Principles of Physics and General College Physics Students: Prob. 1.26: Estimate how long it would take a runner at 10 km / hr to run from New York to California. Explain your solution thoroughly.

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RESPONSE -->

I am little uncertain about where to begin with this problem. I think more information is needed. It would be helpful to know how many hours apart New York is from California are and/or the distance in km.

confidence assessment: 0

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00:23:07

It is about 3000 miles from coast to coast. A km is about .62 mile, so 3000 miles * 1 km / (.62 miles) = 5000 km, approximately.

At 10 km / hr, the time required would be 5000 km / (10 km / hr) = 500 km / (km/hr) = 500 km * (hr / km) = 500 (km / km) * hr = 500 hr.

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RESPONSE -->

ok it is simply hard to solve for a problem if you don't know the needed information.

General knowledge is expected. It is not unreasonable to expect that students know how far it is across the country, and in the absence of this knowledge it is reasonable to expect a student to be able to find it easily.

self critique assessment: 2

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00:32:58

All Students: Estimate the number heartbeats in a lifetime. What assumptions did you make to estimate the number of heartbeats in a human lifetime, and how did you obtain your final result?

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RESPONSE -->

I assumed that the average heart rate for a human being is 72 bpm. I also assumed that the average life span is about 79 years. Therefore,

72 beats * 60 min/hr * 24 hr/day * 365 days/year = 37,843,200 beats

37, 843, 200 beats * 79 years = 2, 989, 612, 800 heart beats in a human lifespan.

confidence assessment: 2

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00:33:33

** Typical assumptions: At 70 heartbeats per minute, with a lifetime of 80 years, we have 70 beats / minute * 60 minutes/hour * 24 hours / day * 365 days / year * 80 years = 3 billion, approximately. **

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RESPONSE -->

close enough! I used the internet to help with these assumptions!

Your answer is more precise than required for a rough estimate. You have a pretty good idea of heart rates and how long people live, and that would have been sufficient. However there is no problem with being more precise than necessary, and your solution is fine.

self critique assessment: 2

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00:34:42

Add comments on any surprises or insights you experienced as a result of this assignment.

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RESPONSE -->

It's hard to answer questions that you have to assume things that not everyone would be approximately correct about.

self critique assessment: 2

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00:34:56

** I had to get a little help from a friend on vectors, but now I think I understand them. They are not as difficult to deal with as I thought. **

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RESPONSE -->

ok

self critique assessment: 3

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