course Phy 201 Would like to have feedback. Thanks, Joshua dyP^assignment #005
......!!!!!!!!...................................
22:54:15 Intro Prob 6 Intro Prob 6 How do you find final velocity and displacement given initial velocity, acceleration and time interval?
......!!!!!!!!...................................
RESPONSE --> 'dt*accel = 'dv 'dv + v0 = vf 'ds = ((vf+v0)/2) *dt confidence assessment: 3
.................................................
......!!!!!!!!...................................
22:54:24 06-30-2008 22:54:24 ** 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**
......!!!!!!!!...................................
NOTES ------->
.......................................................!!!!!!!!...................................
22:56:07 Describe the flow diagram we obtain for the situation in which we know v0, vf and `dt.
......!!!!!!!!...................................
RESPONSE --> from vf and v0 we get 'dv and vAve from vAve and 'dt we get 'ds from 'dv and 'dt we get accel confidence assessment: 3
.................................................
......!!!!!!!!...................................
22:56:13 06-30-2008 22:56:13 ** 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. **
......!!!!!!!!...................................
NOTES ------->
.......................................................!!!!!!!!...................................
22:58:50 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 --> After a little research i determined that from New York to California there is approximately 5000km. 'dt = 'ds/vAve = 5000km/10km/hr = 500hr confidence assessment: 3
.................................................
......!!!!!!!!...................................
22:58:57 06-30-2008 22:58:57 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.
......!!!!!!!!...................................
NOTES ------->
.......................................................!!!!!!!!...................................
23:02:40 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 --> 3.4 billion at a heart rate of 75 bpm living 85 years we have 75bpm * 1440 mpd *365dpy*85y =3350700000 confidence assessment: 3
.................................................
......!!!!!!!!...................................
23:02:46 06-30-2008 23:02:46 ** 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. **
......!!!!!!!!...................................
NOTES ------->
.......................................................!!!!!!!!...................................
23:02:51 06-30-2008 23:02:51 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?
......!!!!!!!!...................................
NOTES ------->
.......................................................!!!!!!!!...................................
23:02:54 06-30-2008 23:02:54 ** 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.. **
......!!!!!!!!...................................
NOTES ------->
.......................................................!!!!!!!!...................................
23:03:18 Add comments on any surprises or insights you experienced as a result of this assignment.
......!!!!!!!!...................................
RESPONSE --> Very informative...learned quite a bit self critique assessment: 3
.................................................
......!!!!!!!!...................................
23:03:28"