assignment 5 query

course phy 231

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005. `query 5

Physics I

02-04-2008

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21:15:49

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 -->

vf=v0+a(time)

confidence assessment: 3

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assignment #005

005. `query 5

Physics I

02-04-2008

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21:27:40

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 -->

Vf=Vo+a(t)

confidence assessment: 3

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21:29:56

Describe the flow diagram we obtain for the situation in which we know v0, vf and `dt.

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RESPONSE -->

the flow diagram would represent the flow of v0, vf,and dt; represented by 3 seperate lines

confidence assessment: 3

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21:32:23

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 -->

would not be able to make such an estimate without knowing the age of the individual

confidence assessment: 3

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21:33:04

** 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 -->

self critique assessment:

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21:33:19

** 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 -->

didnt know what the average lifetime was

self critique assessment: 2

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21:41:02

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 -->

(-2i*6j)+(2i*-3j)=-18

confidence assessment: 2

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21:41:26

** 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 -->

self critique assessment: 0

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21:42:31

** 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 -->

havent done vectors in a while. forgot about squaring abd dealing with magnitudes and so forth

self critique assessment: 1

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You need to include the given solutions and your self-critiques.

In case you didn't get this in the preceding:

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