Lab Report #4

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Lab Report #4

Post  bizk1t4u on Mon Oct 13, 2008 11:10 am

Question 10) The velocity v=d/t . Assume that the error of the time t is negigible (i.e. 1/t is a constant factor for the purpose of error calculation). Given Deltad , the absolute error of the distance d in the figure above, give the expression of the relative error Deltav/v (Use equation (3) of "Error and Uncertainty". Use Deltad for ). Then convert to the expression for the absolute error Deltav (Use :equation (4) of "Error and Uncertainty".)

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Re: Lab Report #4

Post  bizk1t4u on Mon Oct 13, 2008 11:37 am

actually i figured it out, its (Deltad/d)*v

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lab 4

Post  unknown on Tue Oct 14, 2008 12:06 pm


I can't seem to get the units for N and deltah/h, can you help?

also how to do 16)? delta Ke for vn and v1?

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Reponse and a Question

Post  Guest24 on Tue Oct 14, 2008 12:39 pm

For both of those units the answer is "none", because they are unitless.
I have a question also, I was wondering if anyone knew how to find Delta KE and Delta PE...If so a response is greatly appreciated.

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how do u get Q4?

Post  guest12 on Tue Oct 14, 2008 1:48 pm

Enter the expression for the change of the potential energy (see Ch 6 Sheet 12) for the combination of the two Masses M and m when the small mass drops a distance h.(Consider the correct sign of PE!)

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Re: Lab Report #4

Post  unknown on Tue Oct 14, 2008 3:24 pm

Use the lab manual, it ask for change PE in terms of m and h--so use the error manual equation (7)

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ques 7

Post  unknown on Tue Oct 14, 2008 3:44 pm

anyone know what the syntax for ques 7 should look like?

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Question 7

Post  shnickni on Tue Oct 14, 2008 3:54 pm

Delta(Nd)/N

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questions 15 and 16

Post  unknown on Tue Oct 14, 2008 4:07 pm

anyone got up to ques 15 and 16 yet?

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question 16

Post  anan on Tue Oct 14, 2008 4:29 pm

for question 16: .5*(m+M)*(v_n^2-v_1^2)

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question 13, 14, 15, 16, 17

Post  elephant on Tue Oct 14, 2008 5:30 pm

help needed on these question please. thanks
question 13, 14, 15, 16, 17

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thinking hard on these questions

Post  thinking on Tue Oct 14, 2008 5:34 pm

can some help me with question #s
4, 8, 9, 13, 24 and 25
thanks for the help.

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Number 1 & 4

Post  Question on Tue Oct 14, 2008 5:44 pm

For question 1, 4... its keeps asking for the masses combined based on the equations given in our lecture... I'm been playing around with it, but its not making sense:

KE=1/2mv^2 do they mean KE=1/2(m1+m2)v^2??

PLEASE HELP... im sinking

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Lab 4

Post  Guest 87 on Tue Oct 14, 2008 6:46 pm

13.) sqrt((Deltam/m)^2+(Deltah/h)^2) * PE
14.) 2*(Deltav_n/v_n
15.) 2*Deltad/d

Does anyone know! 18,19,20

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Re: Lab Report #4

Post  Guest01 on Tue Oct 14, 2008 6:55 pm

Question wrote:For question 1, 4... its keeps asking for the masses combined based on the equations given in our lecture... I'm been playing around with it, but its not making sense:

KE=1/2mv^2 do they mean KE=1/2(m1+m2)v^2??

PLEASE HELP... im sinking

Question 1: that is exactly what they are asking for. If we go back to the two masses attached by a string over a pulley, we consider the weight of both of them. It is not represented here as m1 and m2, but rather m and M. Remember for 1/2 to either consider it by itself, by enclosing it in parenthesis or to represent it as a decimal.

Question 4: I am having trouble with this one as well, I know the answer, but I do not know how they want it expressed. If PE=mgh and we want to represent the change of it, all we need to do is calculate the change in height, since the height is relative to the PE. This is shown by h1+h0. I believe here is where I'm getting it wrong. I do not know exactly how they want it represented, since you can just call h, "h" or "h initial", "h2-h1". We have to also realize that just as our m in the KE includes both the m and M, so will the m for our PE.

Question 8: These should be values obtained from the experiment. Perhaps you have done the lab, I have not so I am not certain as of yet, but it does seem to have to be a calculated value from the experiment.

Question 9:
Part I: This should look like something similar to our PE from Question 4. Again I do not know what value of h they want entered, but I believe this should look either like something h5-h1 or hn-h1. This should also be multiplied by d.
Part II: They are really asking for the relative error of d. It is found using Equation 3 from Error and Uncertainty. It looks exactly like the relative error of h just with d and the answer is in fact given right there in the question; just write it out in MapleTA syntax.

Question 13: This is the error for PE found using Equation 7 from Error and Uncertainty. Instead of using A and B we are using m and h. Because really the A and B values are just variables, they do not stand for any particular unit called A or B but instead interchangeable with the values and units we currently have. Equation 6 and Equation 7 differ ever so slightly by the fact that (6) is equal to the error, represented by ΔS[i], (7) is the equal to the relative error, not just the error, this is represented by [i]ΔS/S. We know the value of S and are finding it's error ΔS. work it out algebraically so that you are just finding the error (ΔS), not the relative error (ΔS/S). Write both equations out, or look at them and it may make a bit more sense then in your head.

Question 24, 25: These will be calculated values obtained from your experiment. I do not know if you have had lab yet, I have not, but will tomorrow (Wednesday) and will not be able to have the values worked out until tomorrow night or Thursday.

The rest I have not done yet.

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LAB 4 - ANSWER TO #4

Post  phygirl on Tue Oct 14, 2008 7:15 pm

I was having a ton of trouble with number 4 myself and then I realized the mistake... you have to include the sign of mgh. Which is negative. We do not need to alter the equation at all, just plig in the sign.

4.) -(m)*(g)*(h)

YAY!!!

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GRAPH

Post  GUESTT on Tue Oct 14, 2008 7:38 pm

Seeing as how the PE's are negative are we graphing them as being in the negative y-axis???

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#19 & 20

Post  lanthony on Tue Oct 14, 2008 8:46 pm

Does anyone have any insight or help they could offer with 19/20?

I am completely lost on these errors, getting from v^2 to the KE changes between n and 2.

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Graph

Post  B on Tue Oct 14, 2008 9:06 pm

I have the answer to 19 and 20.

First, how does the graph work, since the PE is negative?

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Re: Lab Report #4

Post  lanthony on Tue Oct 14, 2008 9:10 pm

the pe is decreasing so you will have a negative graph going across

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Re: Lab Report #4

Post  B on Tue Oct 14, 2008 9:23 pm

Ok, but how will it be placed on the y axis? In other words, are you saying as the KE increases, the PE decreases? Do we treat the PE as positive? Just putting a negative sign?

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Re: Lab Report #4

Post  lanthony on Tue Oct 14, 2008 9:25 pm

your just going to label the yaxis with negative numbers, yes as the PE decreases, the KE increases

it would be done under the x-axis so all of your y's are negatve. quadrant 3/4 in a graph.

make sense?

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19 and 20

Post  physicss on Tue Oct 14, 2008 9:30 pm

Does anyone know how to do 19, 20?

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Re: Lab Report #4

Post  B on Tue Oct 14, 2008 9:31 pm

I think you mean quadrant 4, where, if you were the draw the 4 quadrant graph, it would be the bottom right one, where only y is negative, and x is positive?

In that case, it makes sense to label the x (KE) number on the top of the graph right?

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question 4

Post  guest007 on Tue Oct 14, 2008 10:19 pm

-m*g*h


hope that works Surprised)

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Re: Lab Report #4

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