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While = camping (np, would Iike camt walcr (TOm Ihc lakc his Campsite Mils two ~identical buckets with water and allaches them [0 Ioog Since the buckcts are MUt ide...

Question

While = camping (np, would Iike camt walcr (TOm Ihc lakc his Campsite Mils two ~identical buckets with water and allaches them [0 Ioog Since the buckcts are MUt identical, he (rals that te Icxl balances ahout [Xint located length 0.4i5= (rom mhelrxint, choxm Thc weight of Uhe rod and empty buckets ) negligible compand the Weight of Ihe water; and the buckets are IHX shown scile:Which bucket holds muc MalcnbucketThey molal Ihe Sme umouneafunicrbucketHow' much morc watee does Uhe heavier buck

While = camping (np, would Iike camt walcr (TOm Ihc lakc his Campsite Mils two ~identical buckets with water and allaches them [0 Ioog Since the buckcts are MUt identical, he (rals that te Icxl balances ahout [Xint located length 0.4i5= (rom mhelrxint, choxm Thc weight of Uhe rod and empty buckets ) negligible compand the Weight of Ihe water; and the buckets are IHX shown scile: Which bucket holds muc Malcn bucket They molal Ihe Sme umouneafunicr bucket How' much morc watee does Uhe heavier bucke contnin, expressed prentage? percenage



Answers

Tor the cellar of a new house, a hole is dug in the ground,
-with vertical sides going down $2.40 \mathrm{m}$. A concrete foundamall is built all the way across the $9.60-\mathrm{m}$ width of the excavation. This foundation wall is $0.183 \mathrm{m}$ away from the front of the cellar hole. During a rainstorm, drainage from the strect fills up the spacc in front of the concrete wall, but not the cellar behind the wall. The water does not soak into the clay soil. Find the force the water causes on the foundation wall. For comparison, the weight of the water is given by $2.40 \mathrm{m} \times 9.60 \mathrm{m} \times 0.183 \mathrm{m} \times 1000 \mathrm{kg} / \mathrm{m}^{3} \times$
$9.80 \mathrm{m} / \mathrm{s}^{2}=41.3 \mathrm{kN}$

In this question. Uh, we have this situation. So a packet eyes, please, on the skill and that water is slowing. Ease. Uh, and it reads off 0.25 Leaders close again. Uh, water buckets, T then, um and this water comes from the heights off 2.6 meters and then the buckets. It's given to be 0.75 kg. Okay, so in this question, we want to find the reading on the skill at three seconds after water starts to accumulate in it. OK, so at three seconds Uh huh. Amounts of water in buckets. If you go to 0.25 times three, the 0.75 leaders. So the mass of water is the density of water. The times the volume, which is one, uh, grand Percy and Q times 0.75 Uh, leaders times, huh? Tens of thousands. Okay, so Okay, this gives us our zero. So her and 50 grams, the Chiefs, I 0.75 kitschy. Okay, then we wants to find the change in momentum of water. Okay. The next thing is want to find a changing moment of my order. Que the reason why we need to do this is that, Ah, as the water flows down and read out was fishing. Um uh, water would experience a change in momentum from the rockets. And so the water would accept a force on the bucket, huh? And that contributes to the reading on the skill escape here to find it out appear water. So first thing to find a finance, speak off water before impact just before impact. Okay, so this is equal to two GH spirit. Okay. And, uh, two times 9.8 times, 2.6 skirts. And this gives us 7.14 meters called second to. So to get this expression eyes by conservation off energy. Okay, um, using the water. Earth sister. Hey, Kate. Then to find her momentum a water changing moment on my water in each second. Hey, this isn't good to, uh, pf minus P I. And then, uh, P Final zero. Okay, so is zero minus p. I Okay, and then I'm going to take downward to be positive. Okay, so in each 2nd 0.25 kids, your water slows down, and then and, uh, this is the finest feet off the water and don what is taken to be positive. So the dot appear water is going to be negative. 1.78 k g. You guys got second. And this so negative sign means that the death appear. Water is, uh, pointing up with. Okay, so the fourth, that impulsive force. Yes, water exits on buckets is equal to, uh, 1.78 new tenants come with. Okay, guy Newt instead. So lastly, we are able to find a reading on the skill que the reading on the scale would be the weight. Oh, buckets. Class the wheat. Oh, water. Plus the impulsive force. Uh, water exits on buckets. Okay, So you have 0.75 No, this is a mess off the packets. Times 9.8. And the weight off water is gain at treats against is that 0.75 kg and multiplied by 9.8 again and plus 1.78 Which is the impulsive force. The water exits on buckets. You at this up, you get 16.5 your kids. So this is the answer for this question.

Question 58 is got a couple things going on. First, we've got a bucket on. A scale on this scale is going to record the forces that are acting on it and the bucket has a massive 580.54 kilograms and then we pour sand into this bucket at a constant rate of 56 grams per second. 56 grams per second is, of course, the same things 560.56 kilograms per second and that sand hits the bucket. It's moving with a velocity of 3.2 meters per second, which means the force generated by that falling sand is 0.56 times 3.2 and that's 0.56 times 3.2 is 0.1792 new moms, Um, what is the reading of this scale when there is 0.75 kilograms of sand in the bucket and what is the weight of the bucket and the 0.75 kilograms of sand? So answer A is different than answer B, because answer is with the scale is reading which the scale has to oppose the four student gravity which is down. But it also has to oppose the force from the falling sand. So the force from the scale actually sums to the force from the sand and the force from gravity. Well, for part A. It says the reading of the scale when there's 0.75 kilograms of sand in the buckets, we have our 0.75 kilograms of sand close. Our 0.54 kilograms of bucket times 9.8 point 75 plus 750.54 times 9.8 is 12.642 However, the scale is also pushing up against the sand, which is exerting a force of 0.1792 so plus 0.1792 and we see that the scale is actually going to have to exert 12.82 Newtons of force to stop the sand and hold it there. The of course, actual weight of this stuff is only 12.64 news at that time

So for this one, it says the laundry's service is buying detergent and fabric from its survivors, Hoffner from its supplier. The supplier will deliver no more than three hundred pounds and a shipment. So, first of all, if it says no more than three hundred pounds, that means three hundred pounds is the maximum. So we can say that the the total total to be less than or equal to three hundred so it could be three hundred because through hundreds of max, but there can't be any more than three hundred. No. So then it says each detergent is seven point three pounds weaken, say, seven point three five d, and each some public softer is six point two pounds, so I can say six point two s, and we know that that total is again that's going to be less than their ego to three hundred pounds. So that's we've got an equation for the wait. We still have to have a number on an equation for the number of boxes or containers I can't in order to get that, we know that the surface wants to buy at least twice as many containers of detergent as containers of fabric softener. So we have to find a way to rip relate the d and the S. If there's going to be twice as many detergents, that means we can say to us because, for example, if we had ten detergents, we know that we have five softener. So if we put that into the equation D is less than two. Us should work. So don't accidentally put the two with the D. C to D s a mess. Because then, if we had that situation two times time, being less than or equal to or greater than or equal to five is not the equation we want to. Some people just see that twice is close to detergents and register to T where we have to think about the equation a little bit more so we know our equation and specially to be D is greater than or equal to us. So we know with this equation and the situation appear or correct, answer should be, eh?

Yeah this question covers the concept of the pressure and the pressure by elemental forces. P. Equals T. F. I. D. We are going to use this equation and let's assume elementary area at a height of edge from the top Of the foundation one and the height of this elementary area is D. H. So we can write the equals the words into th. And uh fourth defund that wall is a. Q. Voluntary. The pressure and the pressure is ah yeah the pressure due to water is roll into G. Into H. Okay into the elementary area that is G. Or we can try it. W. Into th Are the 4th DF equals row g. due to W. H. B. H. Now we can integrate and find the total force exerted by the water on the foundation world. R. F. Equals bro G. W. The grill of H. D. H. From H. Equals 02 H. Equals capital H. Are the total force exerted by the bottles row G. W. Into each square upon. Now, we can substitute the values to find the total force exerted by the water and the force equals the density of water is 1000 kg wow. For me to kill into G. That is 9.8 meters for a second square into the victim and the winters. Ah 9.9.6 m into the square of the height and the height is 2.4 m squared upon to the force exerted by the waters. Uh, 271 into 10 days, three note. Or we can write the forces that you were into 2 71 kg note. If you compare this force compared to the weight of the water, So the force upon the weight is equivalent to 71 kN upon 41.3 generator. The force exerted by the water and the foundation bar is 6.56 times the weight of the water.


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