query_assign5

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20:19:32 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 --> We need to solve for one of the missing variables before we can find the other. Because we are given v0, a and `dt, we can solve for vf. vf = v0+a*`dt From solving for vf, we can then find `ds from `ds = (v0+vf)/2*`dt

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20:20:15 ** 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. There are many ways to do this. I was using the equations in order to become used to them.

It's clear that you have very good mastery of these procedures.

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20:34:00 Describe the flow diagram we obtain for the situation in which we know v0, vf and `dt.

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RESPONSE --> The top layer consists of v0, vf and `dt the second layer could do a few things, my option had vAve, from (vf+v0)/2, assuming uniform acceleration, the next layer had `dv from vAve and v0, as well as `ds from vAve and `dt. The last layer had a from `dv and `dt. It seems this could have variations.

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20:34:21 ** 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

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20:51:12 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 read this problem in the book and also the section on order of magnitude rapid estimating. Can you explain this further with more examples. The examples given, and also their answers are not making sense to me. Here is my attempt. I am estimating approximately 3000 miles from NY to CA. or 3 x10^3. would be divided by the rate of 10km/hr, or approximately 6 mi/hr. My answer would be to divide the 6mi/hr into the 3000 miles, getting approximately 500 hours.

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20:53:27 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 looks similar, except I kept the conversion in miles as opposed to km.

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20:57:21 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 estimated a resting heartbeat to be about 65 beats/minute, and then added in for exercise, sleep, it will vary at these times, so gave an average heart rate of 75 beats/min. From Table 1-2 in the book, an average person lives about 70 years, which is 2x10^9seconds. So, at 75 beats/min converted into seconds (75 beats/min)*(1min/60s) = 1.25 beats/seconds * 2x10^9s = approximately 3 x 10^9 beats in an average lifetime, or 3 billion.

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20:57:53 ** 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.

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20:58:41 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 --> n/a

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20:58:46 ** 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 --> n/a

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20:58:59 Add comments on any surprises or insights you experienced as a result of this assignment.

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RESPONSE --> none at this time

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20:59:04 ** 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

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You did very well on these questions. Let me know if you have questions.