course Phy 231 M????????x?assignment #005005. `query 5
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19:22:31 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 = a * 'dt + v0, which is derived from the formula for acceleration: a = (vf - v0) / 'dt, solved from vf. once vf is known, we can find 'ds by using the formula: 'ds = 'dt * [ (vf - v0) / 2 ], which is derived from the formula for finding vAve = 'ds / 'dt, where vAve can also be stated as: (vf - v0) / 2. confidence assessment: 3
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19:23:24 ** 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 self critique assessment: 3
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19:36:00 Describe the flow diagram we obtain for the situation in which we know v0, vf and `dt.
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RESPONSE --> If v0, vf and 'dt are known, then the flow diagram would look something like: v0, vf, and 'dt on the top line in that order lines from v0 and vf down to the symbol delta-v to represent our reasoning in determining change in velocity. Another line down from delta-v to a and from 'dt (on the top line) to a, further indicates our reasoning of acceleration from the change in velocity and time interval. confidence assessment: 2
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19:37:25 ** 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 - I didn't know you could have two symbols flowing from the same two points - ex: 'dv and vAve from v0 and vf. Had I done this, I would have come up with additional flows. self critique assessment: 2
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19:37:41 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 - not my class. confidence assessment: 3
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19:37:49 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 - not my class self critique assessment: 3
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19:41:52 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 used a resting heartrate of about 80 beats per minute for an adult. I also estimated 75 years, which I believe is roughly the average lifespan of an adult. At 80 beats/minute, 60 minutes/hour, 24 hrs/day, 365 days/year, for 75 years, the total comes to: 3,153,600,000 - or just over 3.1 billion times. confidence assessment: 2
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19:42:11 ** 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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20:01:38 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 --> To find the angle between two vectors, we can use the formula: ( AxBx + AyBy ) / [ ||A|| ||B|| ] = cos(phi) A = < -2, 6 > B = < 2, -3 > [ (-2)*(2) + (6)*(3) ] / [ sqrt( 4 + 36 ) * sqrt( 4 + 9 ) ] = cos(phi) [ -4 + 18 ] / [ sqrt(40) * sqrt(13) ] = cos(phi) 14 / [ sqrt(520) ] = cos(phi) 14 / 22.804 = cos(phi) 0.614 = cos(phi) phi = 52.125 degrees confidence assessment: 3
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20:04:30 ** 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 --> Interesting. I used formula 1.21 from the text but seem to have come up with the wrong answer. I correctly calculated the magnitudes of A and B. Oh, I see! I messed up the negative sign on calculating the dot product. I had -4 + 18, rather than -4 -18, which would have given me the correct number in the numerator. Had I done that correctly, I would correctly identified the answer. self critique assessment: 2
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20:06:24 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I found the equations presented in the Memorize This, Idea 5 to be challenging at first, but then was able to master them sufficiently. I still struggle a little bit with the flow diagrams, but I'm getting better. At first, it didn't seem like a great way to help facilitate finding solutions, but now I see the application a little better. self critique assessment: 3
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20:06:35 ** 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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