course Phy201 h?~???????????assignment #005005. `query 5
......!!!!!!!!...................................
09:52:06 Intro Prob 6 Intro Prob 6 How do you find final velocity and displacement given initial velocity, acceleration and time interval?
......!!!!!!!!...................................
RESPONSE --> vf = 2*vAve - v0 `ds = vAve * `dt confidence assessment: 3
.................................................
......!!!!!!!!...................................
09:53:43 ** 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**
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
10:05:01 Describe the flow diagram we obtain for the situation in which we know v0, vf and `dt.
......!!!!!!!!...................................
RESPONSE --> with v0, vf, and `dt on the top level. v0 and vf connect to `dv on the second level. `dt and `dv connect to a on the second level. v0 and vf connect to vAve on the top level, vAve and `dt connect to `ds on the top level. confidence assessment: 2
.................................................
......!!!!!!!!...................................
10:06:26 ** 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. **
......!!!!!!!!...................................
RESPONSE --> That is exactly what I have, I just don't know which level to label them.
.................................................
......!!!!!!!!...................................
15:38:44 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.
......!!!!!!!!...................................
RESPONSE --> 1km is approximately .6miles The distance from NY to CA is approx. 3000 miles therefore, 3000miles * (1hr/6mi) = 500 hours confidence assessment: 3
.................................................
......!!!!!!!!...................................
15:39:25 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.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
15:47:27 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?
......!!!!!!!!...................................
RESPONSE --> The typical resting heart rate is 70beats/min with a average lifespan of 70 years which is approximately 2*10^9 seconds. Assuming that you are always resting the the total number of beats would be 70beats/min * (1min/60s) * 2*10^9 s => 2.3 *10^9 beats in a lifetime or 2,333,333,333 beats. confidence assessment: 3
.................................................
......!!!!!!!!...................................
15:47:44 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
15:47:59 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?
......!!!!!!!!...................................
RESPONSE --> n/a confidence assessment: 0
.................................................
......!!!!!!!!...................................
15:48:08 ** 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.. **
......!!!!!!!!...................................
RESPONSE --> n/a self critique assessment: 0
.................................................
......!!!!!!!!...................................
15:48:23 Add comments on any surprises or insights you experienced as a result of this assignment.
......!!!!!!!!...................................
RESPONSE --> none self critique assessment: 3
.................................................
......!!!!!!!!...................................
15:48:33 ** 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. **
......!!!!!!!!...................................
RESPONSE --> n/a self critique assessment: 3
.................................................