course Phy 121 ?T?????????? ?z?assignment #????n???????????Physics I
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11:11:37 Intro Prob 6 Intro Prob 6 How do you find final velocity and displacement given initial velocity, acceleration and time interval?
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RESPONSE --> Multiply the a by the `dt to get the `dv. Add the `dv to the v0 to get the vf. (vf - vo) /2 = vAve Finally, `ds is calculated by multiplying the vAve by the `dt
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11:11:47 ** 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**
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RESPONSE --> ok
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11:14:29 Describe the flow diagram we obtain for the situation in which we know v0, vf and `dt.
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RESPONSE --> v0 and vf connect to both the vAve and the `dv. vAve and `dt connect to the `ds. `dv and`dt connect to the a.
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11:14:45 ** 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. **
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RESPONSE --> ok
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11:23:55 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.
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RESPONSE --> Assuming the distance is around 9600km, we would write it in scientific noation as 9.6 * 10^3. Dividing by 10km/hr, we would get : 9.6 * 10^2 hours
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11:25:54 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.
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RESPONSE --> Oh, I guess I didn't measure the scale on my atlas very well. In any case, I do beleive I get this, althoug I thought we were supposed to use order of magnitude estimating.
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11:32:50 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?
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RESPONSE --> My assumptions are that the average heart beats 100 beats per minute and that the the average lifetime is 70 years. which would mean 3.679 * 10^7 minutes times 1 1*10^2 heartbeats= 3.679 * 10^9
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11:32:57 ** Typical assumptions: At 70 heartbeats per minute, with a lifetime of 80 years, we have 70 beats / minute * 60 minutes/hour * 24 hours / day * 365 days / year * 80 years = 3 billion, approximately. **
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RESPONSE --> ok
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11:33:01 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?
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RESPONSE -->
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11:33:04 ** 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.. **
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11:33:06 Add comments on any surprises or insights you experienced as a result of this assignment.
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11:33:09 ** 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. **
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RESPONSE -->
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