course Phy 121 FlҾzƏWwassignment #005
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17:02:15 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 --> When given v0, a, and 'dt, you can find the 'dv by multiplying a * 'dt. The change in velocity added to the initial velocity will give the final velocity. The velocity average can be found by adding the initial and final velocities together and dividing them by 2. Since you now know the velocity average and you already know the time, you can find the displacement by multiplying them together. confidence assessment: 2
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17:02:35 ** 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: 2
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17:07:31 Describe the flow diagram we obtain for the situation in which we know v0, vf and `dt.
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RESPONSE --> v0 vf 'dt 'dv = vf - v0 vAve = (vf + v0) / 2 aAve = 'dv / 'dt 'ds = vAve * 'dt confidence assessment: 2
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17:08:24 ** 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 --> That's what I have. self critique assessment: 2
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17:12:42 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 figured it is approximately 2700 miles from New York to California. 1 mile = 1.609 km. New York would then be approximately 4344 km from California. Running at 10km/hr, you would divide the distance by the time, or 4344.3 km / 10km/hr = 434.43 hr. confidence assessment: 2
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17:13:45 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 --> I figured mine differently but the answer is close, did I do it correctly or did I miss something? self critique assessment: 2
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17:23:13 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'm going to assume that someone's heart beats at an average of 75 beats per minute. If that person lives to be 75 years old, he would have 108,000 beats per day (75 b/m * 1440 m/day), 39,420,000 beats per year (108,000 b/d & 365 d/yr) and 2,956,500,000 beats in their lifetime (39,420,000 b/y * 75 yr). Or, 2.9565 X 10^9 confidence assessment: 2
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17:23:31 ** 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 --> yeah... self critique assessment: 2
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17:23:36 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: 2
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17:23:41 ** 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: 2
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17:25:19 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I need help with number 27 in the book about mowing a football field. The average dimensions of the field are 360 ft X 160 ft. The area of that would be 57600 ft^2. Is that a start? If not, how do I start? self critique assessment: 2
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17:25:29 ** 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: 2
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