course PHY 121 9/14 010. `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. ** STUDENT QUESTION:I wasn’t sure what equation to use to find KE the second way. What does Kef stand for? INSTRUCTOR RESPONSE: In general f stands for 'final' and 0 for 'initial'. Just as v0 and vf stand for initial and final velocities, we'll use KEf and KE0 to stand for initial and final kinetic energies. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating: ---- ********************************************* Question: `qGeneral College Physics and Principles of Physics: convert 35 mi/hr to km/hr, m/s and ft/s. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There are 1.61km per mile. So 35m/hr *1.61km/m = 56.4km/hr. 56.4km/hr * 1000m/k * 1hr/60m * 1m/60s = 15.7m/s. 15.7m/s * 3.3ft/m = 51.81 ft/s. confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK – my numbers were a bit different but I think it’s because I didn’t round exactly the way you did or use the same exact conversions. ------------------------------------------------ Self-critique Rating: ---- ********************************************* Question: `qGen phy and prin phy prob 2.16: sports car 95 km/h stops in 6.2 s; find acceleration YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 95km/h * 1000m/1km * 1hr/3600s = 26.4m/s. `dt = 6.2s. `dv = -26.4m/s/ -(26.4m/s)/6.2s = -4.3m/s^2 confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 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) = -4.2 m/s^2. 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'. STUDENT QUESTION: How did we know that the initial velocity was 0? INSTRUCTOR RESPONSE: The initial velocity wasn't 0. The final velocity was 0, because the car came to rest. Summary of what we were given: • Initial velocity is 95 km/hr, or 26.3 m/s. • Final velocity is 0, since the car came to rest. • The velocity makes this change in a time interval of 6.2 seconds. We can easily reason out the result using the definition of acceleration: The acceleration is the rate at which velocity changes with respect to clock time, which by the definition of rate is (change in velocity) / (change in clock time) The change in velocity from the initial 26.3 m/s to the final 0 m/s is -26.3 m/s, so acceleration = change in velocity / change in clock time = -26.3 m/s / (6.2 s) = -4.2 m/s^2. We could also have used the equations of uniformly accelerated motion, with v0 = 26.3 m/s, vf = 0 and `dt = 6.2 seconds. However in this case it is important to understand that the definition of acceleration can be applied directly, with no need of the equations. (solution using equations: 2d equation is vf = v0 + a `dt, which includes our three known quantities; solving for a we get a = (vf - v0) / `dt = (0 m/s - 26.3 m/s) / (6.2 s) = -4.2 m/s^2.) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating: