course Phy 121 010. `query 10
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Given Solution: 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `qGeneral College Physics and Principles of Physics: convert 35 mi/hr to km/hr, m/s and ft/s. Your solution: 35mi/hr=56.3km/h=15.6m/s=51.3ft/s Confidence rating #$&*: 3
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Given Solution: `aWe 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `qGen phy and prin phy prob 2.16: sports car rest to 95 km/h in 6.2 s; find acceleration Your solution: 26.4m/s/6.2=4.26m/s^2 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 95 km/hr = 95,000 m / (3600 sec) = 26.3 m/s. So change in velocity is `dv = 26.3 m/s = 0 m/s = 26.3 m/s. Average acceleration is aAve = `dv / `dt = 26.3 m/s / (6.2 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'. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating #$&* OK "