course Phy 231 ???`?????I????assignment #005005. `query 5
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08:36:24 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, vf^2 = v0^2 + 2ads, 2ads = vf^2 - v0^2, ds = (vf^2 - v0^2) / (2a). confidence assessment: 3
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08:37:02 ** 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 --> I put the eqns for uniform acceleration, I was wrong. self critique assessment: 2
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08:40:22 Describe the flow diagram we obtain for the situation in which we know v0, vf and `dt.
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RESPONSE --> We use vf and v0 to find dv, we use dv and dt to find ave acc confidence assessment: 3
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08:41:28 ** 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 --> We can ds from v0 and vf? but it didn't say it was uniform self critique assessment: 1
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08:42:02 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 --> OK confidence assessment: 3
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08:42:08 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 self critique assessment: 3
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08:45:03 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 approximted that the human heart beat beats once every second. 60 times in a minute 3600 times in an hour, 604800 times in a week, 31449600 times in a year, and for a lifespan of 75 years: 2358720000 heartbeats. confidence assessment: 3
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08:45:18 ** 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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09:04:41 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 --> 165 degrees confidence assessment: 2
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09:04:57 ** 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 --> OK self critique assessment: 3
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09:05:23 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> No comments self critique assessment: 3
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09:05:33 ** 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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