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Siudent sits Omfrcely rotatlng holding dumbbe cach of mass ko {seC flcure belcw) When hls arms Ihe sluclenl rolales wilh an JmuMId Spe0 744 Tau/s The MUMterL irer L...

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Siudent sits Omfrcely rotatlng holding dumbbe cach of mass ko {seC flcure belcw) When hls arms Ihe sluclenl rolales wilh an JmuMId Spe0 744 Tau/s The MUMterL irer Lia Ihe sludenl plus slool 276 posltlon from the rota on (Flgure b},cxtenaec horlzontally (Fiqure a), thc dumbbell: are 1,08 trom thc axls cf rotatlon Assumed Ve conslanl Tliz sludenl pulls Llie dluribje Tnano horiconlal(a) Firiu Lhedlulcr Speed uf Uie sludenl:ords 'frecly rotatlng stocl" mcar chat thcre arcextcrna tcTJucs ap

Siudent sits Om frcely rotatlng holding dumbbe cach of mass ko {seC flcure belcw) When hls arms Ihe sluclenl rolales wilh an JmuMId Spe0 744 Tau/s The MUMterL irer Lia Ihe sludenl plus slool 276 posltlon from the rota on (Flgure b}, cxtenaec horlzontally (Fiqure a), thc dumbbell: are 1,08 trom thc axls cf rotatlon Assumed Ve conslanl Tliz sludenl pulls Llie dluribje Tnano horiconlal (a) Firiu Lhe dlulcr Speed uf Uie sludenl: ords 'frecly rotatlng stocl" mcar chat thcre arc extcrna tcTJucs applicd Jha consenved this Cosc" radls (b) Find thc kinctic cncray thc rotatina system bcfore and aftcr hc pulls thc dumbbells Inward, Kcarore Kolter -



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The clcctronic configurarions of four clemcnts arc givcn as undcr: (I) $1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{5}$ (II) $1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{4}$ (III) $1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{3}$ (IV) $1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{6 \mathrm{i}} 3 \mathrm{~s}^{2} 3 \mathrm{p}^{1}$ Which of rhe following arrangemcnts gives thc corrcct ordcr in rerms of increasing clectroncgariviry of the clcments? (a) $\mathrm{III}<\mathrm{II}<\mathrm{IV}<\mathrm{I}$ (b) $\mathrm{II}>\mathrm{III}>\mathrm{I}>\mathrm{TV}$ (c) $\mathrm{IV}<\mathrm{III}<\Pi<\mathrm{I}$ (d) $I<$ II $<I I I<\Pi V$

Mm. Hi, welcome to question number 42. In this question, you want to know if we have 25 micrograms of c. 21 age, 30 oh two, We want to know how many moles that is and then how many molecules that represents. So to do this, we are going to need the molar mass of this substance. So we're going to add 21 carbons to the molar mass of 30 hydrogen to the molar mass of two oxygen and I get 314 0.45 So the Mueller mass 3 14 0.45 Of course that would be grams per mole. Right, So I'm going to use this information set up a little bit of dimensional analysis here. First thing I need to do is to get my micrograms two g. So I know that there are one times 10 to the sixth micrograms of anything in one g. Now I can use my molar mass. So there are 314.45 g of C. 21. Age 30 oh two in every mole of the substance. Mhm. Notice I put the grams in the denominator so that they will cancel. So the only unit I have remaining now is moles. So I can go ahead and calculate my answer here. And I'm going to round it to two significant figures since 25 only had two and I get eight point oh times 10 to the negative eighth of the C. 21. Age 30 oh two. Okay, the next thing I want to do then, so let me circle that. That is our first answer. That is the number of moles. Eight times 10 to the negative eight moles. Next thing we want to do is figure out how many molecules this is. So starting with my number of moles eight times 10 to the negative eight moles of C. 21 H 30 02 We need to use of God Rose number because we know there are avocados number of particles in one mole of anything. So six point oh 22 times 10 to the 23rd. That is avocados number. Okay, molecules of the C 21 H 30 02 in every mole of the C 21 H 30 02 What? All right, so mobile cancel and I get a number in number of molecules again rounding the two significant figures. I get 4.8 times 10 to the 16th molecules of C 21 age 30 oh two. So that is the answer to the second part of this question. Thank you so much for joining me.

This question now you just It says if one off the test subjects is randomly selected, find the probability that the subject did not use marijuana. Find the probability of the subject that he did not use marijuana. These are the total subjects who did not actually use marijuana. These are truly negative. These are the ones who are negative and who have also tested negative. But these are the ones who are negative but who have falsely been tested positive. So 24 plus 1 54 is 1 78 1 78 by 300. This is the probability that we want. And do we think the result affect the general population rate off subjects who do not use marijuana? I think the result is somewhat low for people who do not use marijuana. By the way, this value turns out to be 0.59 something zero point Faith Night. And this is my answer

The next story acting on Alba zero. So this means that all the forces acting on it would be zero and all the forces that are acting are B. Which is at 5.5 cm. You can use a different value on. Uh huh. Mhm. Mhm. Yeah mm. Yeah. Yeah. Mhm. Yeah. Mhm. We take all of them to the other side and we saw for A B. And there is 557 136. Mhm. Yeah for part B of the question there is no translation relation in both the horizontal vertical direction So the net force is zero. Mhm. Yeah. Yeah. Take this. Yeah. Cool. Yeah sure. Mhm. Yeah. Okay. Yeah. Mhm. That would be our value for where do you go the information Contraction of the force that the elbow joint exerts and the four armies downwards. I'm good. Mhm. Yeah. Mhm.

So we know that too calculated force when the spring is compressed, will be the spring. Constant times. Thie compression. So let's say initially, the spring is in this state. And then after confession, the spring is compressed, and that's the extra amount ofthe distance that the spring traveled. So if we say that initially the spring is in position X one and after compression it's in the position of ex too. Then we can say that F easy hold. Okay, Times X to minus X one. Now we're given the work done due to the compression, which is Etienne. Jules. Now, since we know the work done due to the compression will try toe relate. This worked on with the spring constant and the compression so that we get the force out of it. So what I mean by that is we know that if the spring is an initial position next one and it's been compressed to a position x two, then the amount ofthe work done is it Has Kate X two squared minus half off key X one squared right. And from there, a cz we say that we already know the work done. So we can actually try to solve for key from this expression and using that key valley, we can actually figure out the force. Now, one thing that we should notice here is initially when the spring is un compressed, we can consider X 10 So if X one is zero, then we can get you off. This time on, we can get you this time over here Now from that, if we try to solve for key, we see that cans equal to times W divided by X two xk red Right on DH. Here the question will be f equals K times x two. Now we can see that if this is the positive extraction, then since it's compressed, we can take extras negative. And we know that the compression is 0.2 meters. So in anarchist X two will be zero point negative zero point two meter. We can use this value to figure out K and using that I came alone, we can solve for the force. So let's do that. Let's write that the expression in terms of work than one more time. So we have X two squared Now if we use the values given So we have the jewels for the world turns and for the displacement. It's negative 0.2. But we can get you out of that negative sign because there's a square here. So even if it's negative, it will be a positive overall. So we can write 0.2 meters and then miss where it's so. Finally, the spring constant will be for 1,000 Newton or meter using that cable we can solve for the force, which is Kate Times X two. So the force will be 4,000 Newton for meteor times X to which is negative 0.2 meter. Using that, we see that the forces negative 800 Newtons and we can say that the magnitude off the force it is 800 mutants now. In the second case, the spring is further compressed down toe 0.4 meters because initially the spring was here from the there that it's been complaints too. 0.2. Then it's been further compressed down. Toa let me draw it here around this position, so the distance from here tto The final position will be 0.4 meters right because it's further compressed down 2.0 toe point toe. So we add point, too, with the previously compressed point. Oh, that's why we got 00.4 meters there. But again that the negative so we can do the similar argument here. Now we'll have x one, a zero on extra as negative 0.4 meters. Amazing that relation. We can again figure out the work done, which is have key X two squared minus half off kids one squared and again we can get it up this time because X one is zero. So from there we see that the Wharton is half off. He is 4,000 Newton per meter, so that's 4,000 per meter. And x two is again a sense. It's a square, so we don't have to worry about the negative sign here. We can just take the magnitude of ex, too, and that's 0.4 squared and combining this, we get the work done as 3 20 Jews. All right, so that's the amount of work done to the spring after it's been compressed to a 0.4 meters. But then the additional work Ah, let's call it W ad. So add means Theoden Ishan ofwork. So the addition and work will be the previous the total work, which is 3 20 Jules minus the worked and do tow the spring being compressed toe 200.2 meters. So this is the additional amount of work, which is 2 40 Jules, right? Um, and for the force, it's going to be again f f equals K Times X, where Katie's 4,000 noon per meter and exes negative 0.4 meter. Using that, we see that the forces negative 1,600 mutants. And if you just if we just take the magnitude ofthe earth, that's going to be 1,600 Nunes. So notice one thing. There we. When we compressed the spring to Onley 0.2 meters, we saw that the work done was on ly 80 jewels. But then when it was when the compression was doubled, Or in other words, when the compression was 800.4 meters, we saw that the worked and was 3 20 Jules and ah, that is that That means that we it's not the world and it's not double, but instead it's actually increasing, um, four times because we see that there's a relation of X squared over here. So if the distance is doubled the work and is four times the previous work. So that's why we had four times the work. And when the spring was compressed toe twice the emission compression. Thank you.


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