phy201
Your 'cq_1_08.2' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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A ball is tossed upward with an initial velocity of 25 meters / second. Assume that the acceleration of gravity is 10 m/s^2 downward.
• What will be the velocity of the ball after one second?
answer/question/discussion: ->->->->->->->->->->->-> :
vf= 25m/s+ 10 m/s^2(1s)= 35m/s
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• What will be its velocity at the end of two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
vf= 25m/s+ 10 m/s^2(2s)= 45m/s
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• During the first two seconds, what therefore is its average velocity?
answer/question/discussion: ->->->->->->->->->->->-> :
vAve= ( 25m/s +45m/s)/2= 35m/s
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• How far does it therefore rise in the first two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
`ds= vAve*`dt so `ds= 35m/s * 2s= 70m
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• What will be its velocity at the end of a additional second, and at the end of one more additional second?
answer/question/discussion: ->->->->->->->->->->->-> :
vf@3s= 25m/s+ 10 m/s^2(3s)= 55m/s
vf@4s= 25m/s+ 10 m/s^2(4s)= 65m/s
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• At what instant does the ball reach its maximum height, and how high has it risen by that instant?
answer/question/discussion: ->->->->->->->->->->->-> :
'ds= (vf^2-v0^2)/2a so `ds= ((0m/s)^2- (25m/s) ^2)/2 (-10m/s^2)= -625/ -20= 31.25 m
vf= v0+at so 0= 25m/s+-10m/s^2 (t) = -25m/s=-10m/s^2= 2.5s
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• What is its average velocity for the first four seconds, and how high is it at the end of the fourth second?
answer/question/discussion: ->->->->->->->->->->->-> :
vAve= (65m/s + 25m/s)/2= 45m/s
`ds= 45m/s* 6s= 180m
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• How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
vf@6s= 25m/s+ 10 m/s^2(6s)= 85m/s
vAve= (85m/s +25m/s)/2= 55m/s
55m/s * 6s= 330m
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8.2
A ball is tossed upward at 15 meters / second from a height of 12 meters above the ground. Assume a uniform downward acceleration of 10 m/s^2 (an approximation within 2% of the 9.8 m/s^2 acceleration of gravity).
• How high does it rise and how long does it take to get to its highest point?
answer/question/discussion:
`ds= (vf^2-v0^2)/2a so `ds= (0m/s^2- 15m/s^2)/2(-10)= -225m/s / -20= 11.25 m = highest point
vf= v0+at so 0= 15m/s+ -10m/s^2 * t so -15m/s= -10m/s^2 (t) so t= -15/-10 = 1.5s= time to highest point
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• How fast is it then going when it hits the ground, and how long after the initial toss does it first strike the ground?
answer/question/discussion:
`ds= v0*t +.5at^2 so 0= 15t + .5(-10) t^2 so 0= t( 15+5t) so t= 0s( @ initial position) and t= 3s ( @ final position)
vf= 15m/s + -10m/s^2( 3s)= - 15m/s
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• At what clock time(s) will the speed of the ball be 5 meters / second?
answer/question/discussion:
vf= v0+ at
5m/s= 15m/s +-10 m/s^2 * t
5-15= -10t
-10= -10t so t= 1s
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• At what clock time(s) will the ball be 20 meters above the ground?
• How high will it be at the end of the sixth second?
answer/question/discussion:
s= s0 + v0 *t +.5at^2
20m= 12m +15m/s *t + .5(-10)t^2
8m= 15m/s *t + -5t^2
0= -5t^2 +15t-8 then using the quadratic equation:
t= -15 +-sqrt( 15^2- 4 (-5)(-8))/ 2(-5)
t= .694s as ball goes up or 2.31s when it comes down
@6s. height(s) =
s= s0 + v0 *t +.5at^2
12m + 15m/s *6sec. + .5( -10m/s^2) * 6^2
s= -78m
•
#$&*
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1 hour
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2.011 s
1.648 s
1.508 s
1.914 s
1.496 s
1.434 s
1.715 s
1.637 s
1.559 s
1.602s
this data is the time it takes for the ball to roll down the incline with one domino placed under the shelf standard and the total distance the ball is rolling from release until it strikes the bracket is 25.0 cm
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1.039 s
1.094 s
1.289 s
1.016 s
.961 s
1.055s
1.032 s
1.055 s
1.047 s
.977s
this data is the time it takes for the ball to roll down the incline with two dominoes placed under the shelf standard and the total distance the ball is rolling from release until it strikes the bracket is 25.0 cm
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.871 s
.832 s
.898 s
.738 s
.789 s
.863 s
.727 s
.730 s
.816 s
.848s
this data is the time it takes for the ball to roll down the incline with three dominoes placed under the shelf standard and the total distance the ball is rolling from release until it strikes the bracket is 25.0 cm
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1.715s., .2359.
1.589s., .1040.
1.079s., .1263.
1.033s., 0.03279.
.8256s., 0.06395.
.7968s., 0.06463.
These results are the mean and standard deviations respectively for left-to-right dominoes followed on the next line by right to left dominoes for 1,2,and 3 dominoes
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16.3265cm/s
17.6211cm/s
25.9400cm/s
27.1055cm/s
33.9147cm/s
35.1406cm/s
These are the average velocities for1 domino, right-to-left, 1 domino, left-to-right, 2 dominoes, right-to-left, 2 dominoes, left-to-right, 3 dominoes, right-to-left, average ball velocity for 3 dominoes, left-to-right
The first line for example is the average velocity for a ball to roll down a ramp from right to left of 28cm with a mean time of 1.715sec. Knowing vAve=ds/dt vAve=28cm/1.715sec=16.3265cm/s (*note that my actual distance was 25cm but 28cm will be used because that is what was instructed)
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/p>9.5198cm/s/s
11.0894cm/s/s
24.0408cm/s/s
26.2396cm/s/s
41.0789cm/s/s
44.1022cm/s/s
These are the accelerations for1 domino, right-to-left, 1 domino, left-to-right, 2 dominoes, right-to-left, 2 dominoes, left-to-right, 3 dominoes, right-to-left, average ball velocity for 3 dominoes, left-to-right
The first line for example is the average acceleration for a ball to roll down a ramp from right to left of 28cm with a mean time of 1.715sec. Knowing a=vAve/ds and that vAve=ds/dt vAve=16.3265cm/s and the time is the mean time for the first trial which is1.715s then a=16.3265cm/s/1.715s=9.5198cm/s/s
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10.3046cm/s/s
25.1402cm/s/s
42.5906cm/s/s
The first line is the average of the accelerations for the right-left setup and the left-right setup using one domino , the second line is the same with 2 dominoes, and the third line is with 3 dominoes, these results are a more accurate acceleration for the ramp with 1 domino because we have results of the average of left-right and right-left t imes.
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16.9738cm/s/1.652s=10.2747cm/s/s
26.5228cm/s/1.056s=25.1163cm/s/s
34.5277cm/s/=.8112s=42.5637cm/s/s
These results are the accelerations using the averages of average velocity with respect to time using the average mean this time for the trials of left-right and right-left setup for each domino trial using 1,2, and 3 dominoes. These results are just more accurate accelerations for each trial
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You would expect the two results to be very close but not exact
They are not the same but they are very close within .03 of each other
You would expect them to be the same because even though they are not done using the same numbers the numbers used to calculate were very similar for instance calculating the average acceleration for left-right and right-left and then averaging them is going to be very close to calculating the acceleration using the mean time instead of the times for left-right and right left
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.03, 10.3046cm/s/s
.06, 25.1402cm/s/s
.09, 42.5906cm/s/s
The left units are the ramp slopes for the 1,2, & 3 domino systems and the right units are the average rate of change of velocity with respect to clock time, ramp slopes are unit less and the acceleration are reported in m/s/s, these results were obtained because ramp slopes for dominoes was given and the (a) was the (a) calculated using change of velocity per rate of time for the left-right and right-left setup for each domino interval and the average for those (a) was used
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.02
-12
No units because this is ramp slope which is unitless
m/s/s because this is (acceleration)
the graph has a y intercept of around -12m/s/s and a x intercept of about .02 there is a best fit line that is increasing
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(.01, -6m/s/s)
I believe the estimate is accurate to the +-.005 slope ramp and to the +-.5 (acceleration)
First number is slope ramp and the second number is (a) in m/s/s these numbers were obtained by estimating based on data from the best fit line
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-.01
-6 m/s/s
600 m/s/s
The slope tells you that just a slight change in one value (slope ramp) can have a significant change on the other value (a)
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1.715s, 1.589s, 1.652s
.2359, .1040, .16995
These results are the means of data from right-left, left- right, and the averages of means of right-left and left right, the standard deviations on the second line follows the same suit
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(left=1.4826, right=1.8225)
The left and right boundaries of the interval represent one standard deviation below the mean and one standard deviation above the mean with the mean being the average of the means and the standard deviation being the average of standard deviations
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18.8857m/s, 12.7383 ms/s/s
15.3635m/s, 8.4299 m/s/s
Min=1.4826s, max=1.8225s
The first line is the velocity and acceleration left-hand boundary
The second line is the velocity and acceleration right hand
the third line is the minimum and maximum possible values of acceleration on this interval
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1.079s, 1.033s, 1.056s
.1263, 0.03279, .079545
1.056s +- .079545
Lower=.9765s, Higher=1.135545s
28.6738m/s, 29.3639 m/s/s
24.6578 m/s, 21.7145 m/s/s
Min=.9765s, Max= 1.135545s
The first line is the mean of the right- left boundary, mean of the left-right boundary, average of the means
The second line is the std. deviation for the right-left boundary, the std. deviation for the left-right boundary, and the average of these std. deviations
The third line is the average of the means and the avg. of the std. deviations
The fourth line is the lower and higher boundaries of the number line
The fifth line is the velocity and acceleration left-hand boundary
The sixth line is the velocity and acceleration right hand
the seventh line is the minimum and maximum possible values of acceleration on this interval
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.8256s, .7968s., .8112s
0.06395., 0.06463., .06429
.8112s +- .06429
Lower= .7469s, Higher=.8755s
37.4883m/s, 50.1918m/s/s
31.9817m/s, 36.5297m/s/s
Min=.7469s, Max=.8755s
The first line is the mean of the right- left boundary, mean of the left-right boundary, average of the means
The second line is the std. deviation for the right-left boundary, the std. deviation for the left-right boundary, and the average of these std. deviations
The third line is the average of the means and the avg. of the std. deviations
The fourth line is the lower and higher boundaries of the number line
The fifth line is the velocity and acceleration left-hand boundary
The sixth line is the velocity and acceleration right hand
the seventh line is the minimum and maximum possible values of acceleration on this interval
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.03, 12.7383m/s/s
.06, 29.3639m/s/s
.09, 50.1918 m/s/s
.03, 8.4299m/s/s
.06, 21.7145m/s/s
.09, 36.5297m/s/s
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My graph does fit this description
The best-fit straight line passes through these three short vertical segments
The best fit line passes through the center of the vertical segments
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Rise=55m/s/s, Run=.09, slope=55m/s/s/.09=611.11m/s/s
These are the rise, run and slope for the coordinates below which are the xintercept and the point on the graph where the steepest line runs through .10 slope ramp
X intercept= (.01, 0) (.10, 55 m/s/s)
Now draw the least-steep possible straight line which passes through all three vertical segments.
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Rise=31.5m/s/s, Run=.10, slope=315m/s/s
These are the rise, run and slope for the coordinates below which are the xintercept and the point on the graph where the steepest line runs through .10 slope ramp
y intercept=(0,1m/s/s) (.10, 32.5m/s/s)
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3 hours
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In the previous lab testing a hypothesis with time intervals do I need to resubmit my lab and if so do I just need to answer the question about whether the hypothesis is correct?
It's easiest just to resubmit a copy of the posted document with insertions marked by ####.
I'm not sure all this is yours.
On the first problem you have given acceleration the same sign as initial velocity. See my notes on your preceding submission, and please resubmit that.