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chcculaleAlu Qalluns CrEOiceu L8v fkxy) = ahere ! Ktr * Tempcrailre ( '( 1 Kh*amclndanor on che around (In Inonc): ddierentiob ebmotette ttunjc In hot chocolate salcs. Roundiour anstenst0 ] decimals rempcncurc ne= Tntng 0{raae Kencher Ince channe hct chorcaate thn Ghanor Omcuint nullnt n Attt Tric; 0/40 chacernan Ent Oeqa n uw [meeruluit K: qullans gt Oid cots chjnac Mne F chocolate sJles &: quilons TaOo 1enerar



Answers

Three odd-shaped blocks of chocolate have the following masses and center-of-mass coordinates: (1) 0.300 kg, (0.200 m, 0.300 m); (2) 0.400 kg, (0.100 m, $-$0.400 m); (3) 0.200 kg, ($-$0.300 m, 0.600 m). Find the coordinates of the center of mass of the system of three chocolate blocks.

We argue in three Masses 0.3 kilograms. 014 kilograms on 0.2 kilograms and they have with the victors. They don't want to come sit upon three meters. You don't want one meter? Come on. Minus 0.4 meter on minus 0.3 meter comma 016 meters. So these are gems and these are excess. Now we know that the center of mass Victor can be found simply by sensational. Am I exciting? Buy some mission or me? Well, except bar are the distance Victor's off each object are the distance vector off the center ofthe mask off the each other. That is, if you have an object with mass m on DH, center of Mass and excite, you can simply put am I excite into this formula and start off substituting for every single object in the system. So we simply write down the formula. HCM bar director is simply the first mass multiplied by the first distance vector that 013 multiplied by zero point comma 013 less 014 But if I by 0.1 come on minus 014 Let's sit a 0.2 multiplied by minus 0131016 divided by 013 Let's see the one for less still want to. Now it's simply this is simply the scale of multiplication. Off Victor's on deuces, they found two multiplied by zero point three is 0.6 Similarly, the second number is 0.79 You go on like that. It was before minus 0.16 Bliss minus 0.6 I was set up on 12 You want a bite? They want to keep a 014712 instead of 19 I was simply out the ex competence and divide it by, say, 19 and we simply rd one components anywhere. 0.9 to get subtle. 0.44 four car Little point 05 56 majors. This is the location Victor off the center of mass off the total system

Using a question. 16.13 and seven didn't want to. You can express the velocity off sound as, ah function off frequency and has a function off the bulk model. It's on dhe. The bulk models can be expressed as negative one over volume. Dying Stevie over Devi with the small piece is pressure on since the volume equivalent and the density row are not changed oppressively so this tree can be taken to the constant on dhe. Then if we take the issue off the frequencies efforts over if I so from this equation, we can talkto because Gloria's over the ice and slammed it is a constant assist. Answers out from the you made it under number Nitto. And then from this a question, we get the same expression to the equal toe V S b s over big guy. And this is true. Also, since stories a constant so that cancels out as well on using this equation. Used this to find B s and P I. So that just becomes equal toe dp over D V s Dover over Dover off dippy over the V I and then they take We scored this and take the inverse off each off the storms. So this will give us Devi over gp s over. Devi over the B I equal Toby I over b s Remember, we have taken the the inverse and also taken are scared or all of this questions. So that is it going to be I will be s on that sick will do if I over F s square because of the door so on were given that f s is 0.333 times if I so if we love that we get the answer Toby almost equal to night.

In this problem on the topic of waves, we are told that if a metal spoon is tapped on a mug of water, the frequency observed his F. I. A spoonful of powder is then added to the water. And the frequency heard as the the mug is tapped, is F. S. And this is due to the change in the waters bulk modular us. Now the bubbles from the water reached the water surface and disappear and the frequency will gradually shift back to its initial value. Now the water's density or volume or these sounds wavelength do not change in this process, but they change the value of the V. P. That is the differential change in volume due to the differential change in pressure caused by the sound wave in water. If we are given the ratio Fs of F. I. To be 0.333 we want to find the ratio DVD ps over D. V. D. P. I. Now we know the speed of sound can be expressed by V, which is the frequency times the wavelength, which is also the square root of the bulk modulates of the material divided by the density of the material. Now we know B is equal to minus G p D V divided by V. Since the lambda and roll are not changed appreciably, the frequency ratio becomes Fs over F I. Is V. S over V. I, which is the square root of the buck module, S B S over B I. Which we can write as the square root of G P Dvs over dP D V I. And so from here, if you rearrange this equation, we get the required ratio deVI Gps divided by T V, G P I is equal to B. I over Bs. And this is equal to the ratio of frequencies if I over F. S all squared. and from the values given this is one over 0.333, all squared. So the required ratio is nine

We can sit that capital and would be the total mass and we can say lower case and would be the melted mass. So given this, we can say that essentially the work divided by the total mass would be equaling than he over row. And so in this case we have then the pressure divided by the density. And we know that the work is contributing to the phase change que Equalling l m where al would be the latent heat of formation for chocolate. Or we can say 150 times 10 to the third jewels per kilogram. And so we find that then the pressure divided by the density multiplied by the total mass equals latent heat of formation multiplied violent, melted ah, mass. And so the melted mass would be equaling toe 5.5 times 10 to the sixth, divided by 1200 this would be 5.5 essentially mega pass cows divided by 1200 kilograms per cubic meter multiplied by mass divided by the late heat of formation 150 times 10 to the third jewels per kilogram. And so we find them that the melted masses equaling two 0.306 where 3.6% of the total mass. We're going to divide this by 0.30 AM And so we divide this by point 30 Um, because again, this would be the mass of the fat. The mass of the fat is eat willing to 30% of the total mass. And so we have that. Then the percentage is equaling 2.0 306 divided by point 30 times 100 percent. And we have This is Equalling 10%. So here what percentage of the fats melt? 10% of the fat melts. That is the end of the solution. Thank you for watching.


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