course Phy 201 åYÕ¶œ[Í‹¸Þòž—ïÚðÝÛ—ýÓÓð÷År§assignment #005
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20:12:35 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 --> Average acceleration is equal to the change in velocity divided by the time interval. You set aAve=(vf-v0)/`dt to get aAve*`dt=vf-v0 aAve*`dt-v0=vf average velocity is equal to the displacement divided by the time interval average velocity is equal to vfinal plus initial velocity divided by 2. Plug in the average velocity and multiply by the time interval to get the displacement. confidence assessment: 2
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20:15:34 ** 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 --> Set aAve=`dv/`dt Multiply aAve by `dt to get `dv Add the change in velocity to the initial velocity to get the final velocity self critique assessment: 2
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20:17:54 Describe the flow diagram we obtain for the situation in which we know v0, vf and `dt.
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RESPONSE --> We know that if we know v0 and vf we can obtain the change in velocity and the average velocity. Because we also know `dt, we can plug in the change in velocity/change in time to get the average acceleration. confidence assessment: 1
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20:19:02 ** 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 --> From the average velocity and the change in time we can also get the displacement. self critique assessment: 2
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20:24:58 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 --> If the person is running at the rate of 10km/hour and runs the distance of about 12000km, we divide the distance by the rate the person runs to obtain 1200 hours. vAve=`ds/`dt vAve*`dt=`ds `ds/vAve=`dt confidence assessment: 1
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20:26:36 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 --> If it is 3000 miles from coast to coast and 1km=.62miles, 3000miles*1km/.62miles=5000 km Then divide the distance traveled by the rate traveled to get the time it took to travel the distance. self critique assessment: 2
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20:31:04 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 --> Assumed: Life expectancy= 80 years 1yr=365 days 1day=24hours 1hr=60min Assumed that if a person has a heart beat of 80 beats/minute and lives for 80yr *365days/1yr*24hours/1day*60min/1hour=42048000 minutes 80beats/minutes*42048000minutes=3363840000beats in a lifetime I multiplied the rate of beats per minute by the number of beats in a lifetime. confidence assessment: 2
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20:31:24 ** 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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20:31:32 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 --> confidence assessment:
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20:31:40 ** 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 --> self critique assessment:
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20:31:46 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> self critique assessment:
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20:31:54 ** 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 --> self critique assessment:
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