course phy 201 assignment #005005. `query 5
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16:27:59 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 --> To find the change in displacement all we have to do is multiply acceleration by the time interval. To find the final velocity we must multiply the acceleration times the time interval and that gives us the change in velocity and then we must take the change in velocity and add the initial velocity to that. confidence assessment: 3
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16:28:18 ** 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 --> I answered this question completly self critique assessment: 3
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16:30:45 Describe the flow diagram we obtain for the situation in which we know v0, vf and `dt.
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RESPONSE --> we first take v0 and vf and add them and divide by two and we have the average velocity and then we can multiply our average velocity by our change in time than that gives us our change in displacement. confidence assessment: 3
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16:32:08 ** 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 I missed the fact that i could have also calculated the acceleration and the change in velocity from the information that was given I will make a note of this mistake and use it later in the semester. self critique assessment: 3
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16:39:58 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 --> well according to mapquest the distance between New York and California is 4,441.5km so if our runner is maintaining an average of 10km per hour then he would be running for 444.15 hours. confidence assessment: 3
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16:41:02 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 --> this is the principal i used to get my answer. we had different milages between the two locations. self critique assessment: 3
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16:45:28 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 --> at 60 beats per minute this is 86,400 beats per day. There are 365 days in a year so this would be an average 31,536,000 beats per year, and using an average lifespan of a male which is 76 years would is 2,396,736,000 beats. confidence assessment: 3
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16:46:10 ** 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 --> I worked my problem the same way my numbers varied from your estimation. self critique assessment: 3
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16:46:45 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 --> I am a general physics student this problem does not apply to me confidence assessment: 3
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16:47:01 ** 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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RESPONSE --> ok self critique assessment: 3
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