course Phy 201 ü‡ñ·fl£«È“˛õ≤•⁄}¬aÄ≈ÓG¨|¢ôxflassignment #005
......!!!!!!!!...................................
20:38:48 Intro Prob 6 Intro Prob 6 How do you find final velocity and displacement given initial velocity, acceleration and time interval?
......!!!!!!!!...................................
RESPONSE --> the final velocity can be found by using the initial velocity added to the acceleration*`dt. displacement is found by multiplying the average velocity by the time interval.
.................................................
......!!!!!!!!...................................
20:38:58 ** 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
.................................................
......!!!!!!!!...................................
20:40:06 Describe the flow diagram we obtain for the situation in which we know v0, vf and `dt.
......!!!!!!!!...................................
RESPONSE --> v0 and vf can be used to find `dv which can be used with `dt to find aAve. then, v0 and vf can be used to find vAve with then is used with `dt to find the `ds.
.................................................
......!!!!!!!!...................................
20:40:11 ** 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 --> ok
.................................................
......!!!!!!!!...................................
20:41:46 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 --> I estimated the width of the US to be approximately 3000 miles. this, when multiplied by 5280 ft/mi, 12 in/ft, 2.54 cm/in, 1m/100cm, and 1km/1000m gives us 4828 km. IF the runner is going at 10km/h that gives a little less than 500 hours to run across the US, assuming they could keep up their speed.
.................................................
......!!!!!!!!...................................
20:41:54 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
.................................................
......!!!!!!!!...................................
20:44:03 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 --> If there is an average of 80 heartbeat per minute and the person lives an average of 70 years, then in that lifetime there will be: 80 hb/min*60min/hr*24hr/day*365days/yr=42048000 heartbeats per year. for 70 years that equals:2.94x10^9 heartbeats in a lifetime.
.................................................
......!!!!!!!!...................................
20:44:21 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 --> ok
.................................................
......!!!!!!!!...................................
20:44:27 ** 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 --> na
.................................................
......!!!!!!!!...................................
20:45:03 Add comments on any surprises or insights you experienced as a result of this assignment.
......!!!!!!!!...................................
RESPONSE --> the uniformly accelerated motion equations are in depth and do not stick in my head as much as i would like them to. I am having trouble memorizing them.
.................................................
......!!!!!!!!...................................
20:45:10 ** 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 --> ok
.................................................
"