Lab Report #3
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Re: Lab Report #3
Your averages are only from the three readings you did from each height.
For Time it will be he same just replace "x" with "t"(x_{1}+x_{2}+x_{3})/3
Guest01 Posts : 133
Join date : 20080919
Re: Lab Report #3
Are you using the underscores and all lowercase? and just the numbers?
I assume this is for x and t and not their errors correct?
I assume this is for x and t and not their errors correct?
Guest01 Posts : 133
Join date : 20080919
#12 and #15
Does anyone know the format for #12? For #15 I'm having trouble finding Deltag I know that i have to use the formula from #13 (sqrt(((Deltah/h)^2)+((2*(Deltak/k))^2)))*g but it's not coming out right. My values are Deltah=.002, h=.9640, Deltak=.56402, k=.4103, g=11.453. Please help!
munchkin Guest
Graph question
Okay, so I graphed my points using distance on the vertical axis and velocity as horizontal distance. I drew my best fit line and found my slope. However, when I plug my slope into the equation g=2h/k^2 I get 64.### . I plotted the distance in meters because the velocity was already in m/s. I found the slope to be k=.1733
my height was 96.4cm = .964 m
g=2(.964)/(.1733^2) = 1.928/.03003289 = 64.196
What am I doing wrong that produces this ridiculous g value?
my height was 96.4cm = .964 m
g=2(.964)/(.1733^2) = 1.928/.03003289 = 64.196
What am I doing wrong that produces this ridiculous g value?
guest45 Guest
Re: Lab Report #3
Munchkin 
Try entering the number when you find it's square, that is before you multiply by your gravity. So leave out your g value.
[I had a bit better of an explanation but the field cleared itself.
Guest45 
What formula are you using to get your slope?
You should be using (Delta y)/(Delta x) and the values that are in your graph, and not the expressed formula.
Try entering the number when you find it's square, that is before you multiply by your gravity. So leave out your g value.
[I had a bit better of an explanation but the field cleared itself.
Guest45 
What formula are you using to get your slope?
You should be using (Delta y)/(Delta x) and the values that are in your graph, and not the expressed formula.
Guest01 Posts : 133
Join date : 20080919
Re: Lab Report #3
yeah i did k=deltay/deltax and i used the points (1.4, .29), (.65, .16) with velocity as the x values in m/s and distance as y values in meters. So then I did k=(.29.16)/(1.4.65) = .1733
I plugged that into g=2h/k^2 with h being 97.4 cm or .974m and i got g=64
I plugged that into g=2h/k^2 with h being 97.4 cm or .974m and i got g=64
Guest 45 Guest
Re: Lab Report #3
Are you using your x_{avg} and v_{avg} values? Did you calculate the values correctly? Have you also completed any of the previous questions correctly? Maybe use the actual formula for k [k=√((2*h)/g)]
When I reverse my formulas for slope I am getting the opposite problem you are.
Using that I get a value of 9.81000001 (give or take a zero), which is pretty dead accurate on the value of g.
When I reverse my formulas for slope I am getting the opposite problem you are.
Using that I get a value of 9.81000001 (give or take a zero), which is pretty dead accurate on the value of g.
Guest01 Posts : 133
Join date : 20080919
Re: Lab Report #3
yeah im using the xavg and vavg values, and maple ta says all my questions are right thus far, including my slope. All I have left is to calculate g. I can't use the k formula without a g. Why does this happen to me
Guest45 Guest
Question 9
Hey, can somebody help me with Q#9? what is deltah? It was given in a class .0002, but its still wrong. I dont know what to put. Please help
guest5 Guest
Re: Lab Report #3
silly question, but when we initially recorded time was it in milliseconds?
confused Guest
Re: Lab Report #3
so then we change the milliseconds to seconds by dividing by 1000, right? and then we do deff in meters/avg time in seconds to get m/s for velocity
but then i get a large velocity of like 234.5, 865.35, 1253.78. this all messes up my g value. my seconds are then like .000167, velocity is 234.5 and i get like 5432684.227 as a g...
but then i get a large velocity of like 234.5, 865.35, 1253.78. this all messes up my g value. my seconds are then like .000167, velocity is 234.5 and i get like 5432684.227 as a g...
confused Guest
Help with table 2 and Slope calculation
I did the calculation for table 2 and maple TA is telling me that my calculation for the Lowest and 2nd point on delta T are incorrect. I checked twice. Could someone give me the formula to calculate it, I may be missing something ( although it is saying my other values are correct).
Also How do you go about calculating delta G after you have figured out G from the formula g= 2h/k^2.
Thanks everybody !
Also How do you go about calculating delta G after you have figured out G from the formula g= 2h/k^2.
Thanks everybody !
BGGirl Posts : 6
Join date : 20080921
Re: Lab Report #3
Guest 45
Perhaps there is still a wrong calculation? Have you tried entering the value for normal g which is 9.81? I thought it asked for that from me at first so both normalg and myg values came up correct.
guest5
The given error was 1.1mm, or 0.0011m. Entering this into Equation 6 probably should come out to 0.001 as your error. I handed my lab in very early so if I remember correctly you had to use Equation 6 for this one.
As for the time, I never did any time conversions, I thought all was recorded in seconds and not milliseconds. This may be why a lot of your calculations are wrong, try not converting ms into s, leave it be, already as s
confused
your velocity is equal to the diameter of the ball divided by the average time of the particular launch height. Try not converting the time.
BGGirl
the error is the difference between the highest and lowest number divided by two. Pay attention in table one, because the three may not be in the correct order. For instance the max may have been your second try, and the slowest was perhaps your third try.
The error of g that you entered into Maple TA should of been (sqrt(((Deltah)/h)^2+((2*Deltak)/k)^2))*g, however when you actually calculate it you do not have to multiply it by the final g, that is the g that you had measured. Basically treat it like Equation 6 instead of Equation 7 from Error and Uncertainty.
I know this all a bit late, I'm sorry, but better late then never.
Perhaps there is still a wrong calculation? Have you tried entering the value for normal g which is 9.81? I thought it asked for that from me at first so both normalg and myg values came up correct.
guest5
The given error was 1.1mm, or 0.0011m. Entering this into Equation 6 probably should come out to 0.001 as your error. I handed my lab in very early so if I remember correctly you had to use Equation 6 for this one.
As for the time, I never did any time conversions, I thought all was recorded in seconds and not milliseconds. This may be why a lot of your calculations are wrong, try not converting ms into s, leave it be, already as s
confused
your velocity is equal to the diameter of the ball divided by the average time of the particular launch height. Try not converting the time.
BGGirl
the error is the difference between the highest and lowest number divided by two. Pay attention in table one, because the three may not be in the correct order. For instance the max may have been your second try, and the slowest was perhaps your third try.
The error of g that you entered into Maple TA should of been (sqrt(((Deltah)/h)^2+((2*Deltak)/k)^2))*g, however when you actually calculate it you do not have to multiply it by the final g, that is the g that you had measured. Basically treat it like Equation 6 instead of Equation 7 from Error and Uncertainty.
I know this all a bit late, I'm sorry, but better late then never.
Guest01 Posts : 133
Join date : 20080919
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