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course Phy 201

IùüœÂ×ΥѻN¦}¤É§P½ÙÈkÛãassignment #005

005. `query 5

Physics I

09-11-2008

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02:45:31

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 -->

vf=v0+a*`dt

`ds=(vf+v0)/2*`dt

confidence assessment: 3

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02:46:00

** 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 -->

0k

self critique assessment: 3

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02:52:11

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=`dv,`dt and `dv on the second level=aAve,(vf+v0)/2=vAve,you take vAve on the second level and * by `dt on first level =`ds

confidence assessment: 3

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02:52:45

** 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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03:13:26

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 -->

Lets assume from NY to California is 3500miles. We need to convert this to km. 1km is equal to .6214miles. 3500*.6214=5632km.The guy runs at 10km/hr. We need to divide the distance by his average speed.5632km/10km/hr=563hrs

confidence assessment: 2

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03:13:53

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 -->

little off

self critique assessment: 3

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03:26:31

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 -->

Lets assume the normal heart rate is 70bpm maybe less when sleeping and more when running. Lets say on average you have a heart rate of 85bpm. An average life span is 70years which is app. 2*10^9s.Since there is 60 secs in a min we have about 1.4bp. multiply that by our life span on earth and it is about 2.8*10^9 or 3*10^9heartbeats

confidence assessment: 3

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03:26:53

** 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 -->

ok

self critique assessment: 3

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03:27:05

University Physics Students Only: Problem 1.52 (i.e., Chapter 1, Problem 52): Angle between -2i+6j and 2i - 3j. What angle did you obtain between the two vectors?

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RESPONSE -->

na

confidence assessment: 0

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03:27:44

** For the given vectors we have

dot product =-2 * 2 + 6 * (-3) = -22

magnitude of first vector = sqrt( (-2)^2 + 6^2) = sqrt(40)

magnitude of second vector = sqrt( 2^2 + (-3)^2 ) = sqrt(13)

Since dot product = magnitude of 1 st vector * magnitude of 2d vector * cos(theta) we have

cos(theta) = dot product / (magnitude of 1 st vector * magnitude of 2d vector) so that

theta = arccos [ dot product / (magnitude of 1 st vector * magnitude of 2d vector) ]

= arccos[ -22 / ( sqrt(40) * sqrt(13) ) ] = arccos ( -.965) = 164 degrees, approx.. **

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RESPONSE -->

na

self critique assessment: 3

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03:29:50

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

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RESPONSE -->

The equations of uniformly accelerated motion was a little confusing.Still not comfortable with it

self critique assessment: 2

You appear to be using the equations correctly, and with the practice you will get in subsequent assignments I expect you'll have no trouble getting more comfortable with them.

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03:30:08

** 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. See my notes and let me know if you have questions. &#