course Phy 121 ??R??e??? ??????assignment #010????n???????????Physics I
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14:06:53 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.
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RESPONSE --> We need to find the acceleration by the formula a = F/m. we take the aceleration and apply it to the time interval and come up with the velocity. Now that we know v, we can use the equation: KE = .5mv^2 we can also find the `ds by finding the vAve and use this with another equation: `dw = F * `ds. the `dw will be our KE
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14:07:09 ** 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. **
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RESPONSE --> ok
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14:08:11 General College Physics and Principles of Physics: prob 2.04 convert 35 mi/hr to km/hr, m/s and ft/s.
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RESPONSE --> 56.3km/h 15.6m/s 4.75ft/s
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14:09:36 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.
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RESPONSE --> OK I messed up the last one because I used the second conversion and I should have gone back to the original.
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14:10:30 Gen phy and prin phy prob 2.16: sports car 95 km/h stops in 6.2 s; find acceleration
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RESPONSE --> The acceleration is 4.26m/s
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14:10:39 ** 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'.
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RESPONSE --> ok
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14:10:44 univ phy 2.66 train 25m/s 200 m behind 15 m/s train, accel at -.1 m/s^2. Will the trains collide and if so where? Describe your graph.
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RESPONSE -->
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