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Given _ relation A0you experimentally measured A0 and and grapheditas Af being on the vertical axiz and on tno horizontal axiz Your objective to cetermine tho valuo...

Question

Given _ relation A0you experimentally measured A0 and and grapheditas Af being on the vertical axiz and on tno horizontal axiz Your objective to cetermine tho valuo and the uncertainy of Assume and Q are constants Youconducted linear ft/regrezsion to the data LINEST) and obtained the slope Whate the exprezsion forslope Lsiopeslepe Lx

Given _ relation A0 you experimentally measured A0 and and grapheditas Af being on the vertical axiz and on tno horizontal axiz Your objective to cetermine tho valuo and the uncertainy of Assume and Q are constants Youconducted linear ft/regrezsion to the data LINEST) and obtained the slope Whate the exprezsion for slope L siope slepe Lx



Answers

Here are the $K_{\alpha}$ wavelengths of a few elements:
$$\begin{array}{lll}{\text { Element }} & {\lambda(\mathrm{pm})} & {\text { Element }} & {\lambda(\mathrm{pm})} \\ \hline \mathrm{Ti} & {275} & {\mathrm{Co}} & {179} \\ {\mathrm{V}} & {250} & {\mathrm{Ni}} & {166}\\ {\mathrm{Cr}} & {229} & {\mathrm{Cu}} & {154}\\ {\mathrm{Mn}} & {210} & {\mathrm{Zn}} & {143}\\ {\mathrm{Fe}} & {193} & {\mathrm{Ga}} & {134}\end{array}$$
Make a Moseley plot (like that in Fig. 40-16) from these data and verify that its slope agrees with the value given for $C$ in Module $40-6 .$

This question we have to find slope A. And the deflection at point D. So um Star A. Is T. B. A. Or al from the um moment area theorem. And TBA is 1/2 mine spl over to E. I Tom L. Tom L. Over three. Right by our and that is P elsewhere over 12. Yeah. So um T. D. A. It is 1/2 times when spl over to E. I. Tom al um L over two plus hours over three plus one or 2 p. L. or two E I. L. Over to el over three. And that is minus P. L. To the power of three, divide by four E. I. And delta D. It is to minus three or 2 hour Dwight by al T B A. And it happened it is P. L. To the power of three, divide by eight E. I. These are the answers.

This question we have to find the slow by point C. And the deflection at point B. So first, if we throw the M. O. V. I diagram and used the moment area theorem, we can find that see the sea A. Is equal to see and that is minus two P. A. What about E. I tom a Plus 1/2 minus P. A. O E I A. From the moment and their parents And it is -5 p. a square over to E. I. And said to be. And all the B. S. T. B. A. is 1/2 timeliness. P. A over E. I. Time a time to a 43 Plus 1/2 times when that's P. A. Or E. I. Tom A Some a plus to a over three Plus -2 P. A. Or E. I. Tom A. Um a plus 802. And that is 25 P. A. To the power three. Well by six e. i. Mhm. In a society. And that is the answer.

This question we have to find the slope A. And the deflection at the point B. So from the uh figure in the textbook uh that is a distributed lower W. X. On the beam A. C. So if we use the method of super position, we have that fight of um W. And S. E. Is equals the effect of the distributed Lord on the session A. B. And the moment that on the Roller derby. So we have to use the uh method of superposition for this one is formed, distribute that lower, This one is formed moment at B. And distributed load here at the section A. B. So this version can be found in the appendix. Now we have that um out here is to A. So have that seat a. one is W. A. to the power of three, divide by three b. i. So as uh be out of the the one that is five W. A. to the power of four. Bye bye. And full E. I. And for six A. Two M. Zero. It is W. A. Square divide by two and now is also 28 So see that a. two is W. A. or three went by six E. I. And but it's one uh Oh daddy too. The X. Square here is a square. So we have that ready to is W. A. to the power of four. Wide by eight e. i. So from the method of work position, see that A. Is stay one plus Suit to A. two. And that is minus W. H. With power three. Bye Bye six e. i. Okay. And the 30 is 31 plus The D. two. And that is W. A. or four right by 12 E. I. These are the answers.

This question we have to find slope at point C and the deflection at point B. So if we use the moment area throwing, you have that CSC is 1/2. I'm minus W a square or E i time a plus -5 Square or two E I I'm A. And that is W A. To the power of three White by E I. And that would be, it is T B A. And that is 1/2 minus W a square or E I time A time A plus to over three a plus -5 Square or two. E I Some a some a plus 802 Plus 1/3 -3 WA Square or two E I tom a three or 4 a. And that is 41 W h. D. Power of four. Right by 24. Uh these are the answers.


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