course Phy 201 ?L?^???D}??oz??????assignment #010
......!!!!!!!!...................................
23:29:11 Query introductory problem set 3 #'s 7-12 Describe two ways to find the KE gain of an object of known mass under the influence of a known force acting for a given time, one way based on finding the distance the object moves and the other on the change in the velocity of the object, and explain why both approaches reach the same conclusion.
......!!!!!!!!...................................
RESPONSE --> The change in KE= .5(m)v^2 and Work=distance *force `dw=change in KE `dw=KEf-KE0 Work is equal to the change in KE confidence assessment: 1
.................................................
......!!!!!!!!...................................
23:30:58 ** First way: KE change is equal to the work done by the net force, which is net force * displacement, or Fnet * `ds. Second way: KE change is also equal to Kef - KE0 = .5 m vf^2 - .5 m v0^2. **
......!!!!!!!!...................................
RESPONSE --> KE change=Work done by the net force=Net force * displacement KE change=KEf-KE0=.5m vf^2-.5mv0^2 self critique assessment: 2
.................................................
......!!!!!!!!...................................
23:38:44 General College Physics and Principles of Physics: prob 2.04 convert 35 mi/hr to km/hr, m/s and ft/s.
......!!!!!!!!...................................
RESPONSE --> 1mi=1.609km 1mi=1609m 1mi=5280ft 35mi/hr *(1609km/mi)=56.315km/hr 35mi/hr *(1609m/1mi)(1hr/3600s)=15.64m/s 35mi/hr *(5280ft/1mi)(1hr/3600s)=51.33ft/s confidence assessment: 2
.................................................
......!!!!!!!!...................................
23:39:34 We need a conversions between miles and meters, km and ft, and we also need conversions between hours and seconds. We know that 1 mile is 5280 ft, and 1 hour is 3600 seconds. We also know that 1 inch is 2.54 cm, and of course 1 foot is 12 inches. 1 mile is therefore 1 mile * 5280 ft / mile = 5280 ft, 5280 ft = 5280 ft * 12 in/ft * 2.54 cm / in = 160934 cm, which is the same as 160934 cm * 1 m / (100 cm) = 1609.34 m, which in turn is the same as 1609.34 m * 1 km / (1000 m) = 1.60934 km. Thus 35 mi / hr = 35 mi / hr * (1.60934 km / 1 mi) = 56 (mi * km / (mi * hr) ) = 56 (mi / mi) * (km / hr) = 56 km / hr. We can in turn convert this result to m / s: 56 km/hr * (1000 m / km) * (1 hr / 3600 sec) = 15.6 (km * m * hr) / (hr * km * sec) = 15.6 (km / km) * (hr / hr) * (m / s) = 15.6 m/s. The original 35 mi/hr can be converted directly to ft / sec: 35 mi/hr * ( 5280 ft / mi) * ( 1 hr / 3600 sec) = 53.33 ft/sec.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
23:45:48 Gen phy and prin phy prob 2.16: sports car 95 km/h stops in 6.2 s; find acceleration
......!!!!!!!!...................................
RESPONSE --> v0=0 vf=95km/hr `dt=6.2s a in m/s^2 aave=`dv/`dt `dv=95km/hr=95km/3600s (95km/3600s)/6.2s= .004256km/s^2 1km=1000m .004256km/s^2*1000m/1km= 4.256m/s^2 confidence assessment: 2
.................................................
......!!!!!!!!...................................
23:49:29 ** 95 km/hr = 95,000 m / (3600 sec) = 26.3 m/s. So change in velocity is `dv = 0 m/s - 26.3 m/s = -26.3 m/s. Average acceleration is aAve = `dv / `dt = -26.3 m/s / (6.2 s) = -4. m/s. So the time to come to a stop is `dt = `ds / vAve = 50 m / (12.5 m/s) = 4 s. Acceleration is rate of velocity change = change in velocity / change in clock time = -25 m/s / (4 s) = -4.2 m/s^2. Extension: One 'g' is the acceleration of gravity, 9.8 m/s^2. So the given acceleration is -4.2m/s^2 / [ (9.8 m/s^2) / 'g' ] = -.43 'g'.
......!!!!!!!!...................................
RESPONSE --> IF vf=0 then `dv=0-26.3m/s=-26.3m/s aave=`dv/`dt=-26.3m/s/6.2s=-4m/s^2 self critique assessment: 2
.................................................