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The figure below shows section of hot-water pipe consisting three parts: straight; horizonta plece 28.0 cm long_ stralght vertical piece { = 110 cm long- and a elbo...

Question

The figure below shows section of hot-water pipe consisting three parts: straight; horizonta plece 28.0 cm long_ stralght vertical piece { = 110 cm long- and a elbow Joint that allows the two pipe sectons meet at right angle:stud and second-story (loorboard hold the ends of this section pipe stationary: The plpe made of = copper alloy wlth coefficient of linear expansion of 7 # 10 -S('c When thc water Ilaw Is turned on_ the temperature of the pipe rises from 18.0*C to 42.5*C. What the magn

The figure below shows section of hot-water pipe consisting three parts: straight; horizonta plece 28.0 cm long_ stralght vertical piece { = 110 cm long- and a elbow Joint that allows the two pipe sectons meet at right angle: stud and second-story (loorboard hold the ends of this section pipe stationary: The plpe made of = copper alloy wlth coefficient of linear expansion of 7 # 10 -S('c When thc water Ilaw Is turned on_ the temperature of the pipe rises from 18.0*C to 42.5*C. What the magnitude and direction of the displacement of the pipe elbow from Its original location once the pipe reaches the higher temperature? (Give the magnitude mm and the direction degrees below the tX-axis_ which is parallel t0 the horizontal piece und points from the stud to the clbow Joint:) magnitude dlrection H



Answers

Inside the wall of a house, an L-shaped section of hot-water pipe consists of three parts: a straight horizontal piece $h=28.0 \mathrm{cm}$ long, an elbow, and a straight, vertical piece $\ell=134 \mathrm{cm}$ long (Fig. $\mathrm{P} 10.51 )$ . A stud and a second- story floorboard hold the ends of this section of copper pipe stationary. Find section of copper pipe stationary. Find the magnitude and direction of the displacement of the pipe elbow when the water flow is turned on, raising the temperature of the pipe from $18.0^{\circ} \mathrm{C}$ to $46.5^{\circ} \mathrm{C}$

Hello and welcome to Chapter 16. Problem number 16 in principles of physics. So this problem tells us that we have two pieces of wood to form a junction here with an elbow at origin. Oh, on they follow our coordinates. Positive. Why? With R l positive expert H. It also tells us that we're gonna have ah, thermal expansion of this conjunction here when water pipe is turned on on giving us a temperature change a swell. So I have our thermal expansion equation here. Delta l is Alfa Times. L not times Delta t I as well have the thermal expansion coefficient for copper. That's the material that would be expanding. Hearing problem eso First things first. We need Teoh. We're gonna be first calculating our change in both the horizontal and vertical parts. That's gonna be the total as its of the problem asks us, we need to find that total displacement here, so we need to calculate our our change and lengths. So we first need to get that delta T temperature. Just get that is a number. And then we can, um, find our individual delta elves for both vertical and horizontal components. So our Delta T in the beginning is just gonna be that 45 point are 46.5 minus 18. So we'll have about 28.5 degrees of change here. Now that we have that we can pretty much just plug in our individual values for or l knots And then are coefficient of thermal expansion for copper as well That Delta team. So the delta l for our vertical component, it's gonna be 17 times in the line of sex times at 0.28 Remember, we also have to do We'll have to convert our our lengths. And commuters, I have that up here. They give us, give it to us and centimeters Regardless, we have 17. Excuse me, 17 Thompson of my six times 60.28 times at 28.5 and we'll have a verdict component. A change of 1.35665 cent alliance for and the very same process for our horizontal component. You just have ah, 1.34 as are all not in this equation here. So we have both of our components here. But again, these air just the Delta Els. I'm gonna go back up toward diagram here. This is just the Delta L, um, And for both of these individually, it asks us the total displacement and direction of this elbow at origin. Oh, here. So what we're gonna actually get to do next is used the distance formula to find that magnitude. Eso, if you recall distance formula is gonna be one component of the of our of our direction. Change squared, plus the other components square. Take the square root of that. So, um, we're gonna have d when we plug in our numbers. Here. Be about 6.63 times 10 to the minus, four meters on. Just again. Reminder. Um, this is Delta L l 1.3566 times in the lines. Four squared plus 6.4923% of the minus four squared and then take the square root of that. So that's our distance here on now. In order to find the direction we're gonna look at Delta l, uh, of the vertical and horizontal components as part of a triangle is well, and then we're gonna use trigonometry. I chose this data here. You can choose the angle up here is Well, uh, he or again, it doesn't really matter which angle uses lines you have give that the sense of direction here. So I have, um, using, you know, pretty standard trigonometry rules we have there's equal toward inverse tangent are arc tangent of the opposite toe. Uh, if you remember, so Cotto of the opposite component l vertical over our horizontal are adjacent component l horizontal. So inverse tangent of delta L for the vertical and then divided by Delta off of the horizontal. And again, when we plug those numbers in, we're going to get about 78.20 degrees.

First things you want to do is look up the average linear expansion coefficient for these copper pipes in the textbook, so that Alfa is going to be equal to 17 times 10 to the negative six. That's gonna be enough. Inverse decrease Celsius and we can calculate Delta T but attracting t I from TF. And that's gonna be equal Teoh 28.5 degrees Celsius. Now to find how much it's section of the pipe expands we can use the equation told to l is equal to health food times initial length. I'm still titty and substitute in H and fell lower case l set of our Bigelow and ally. So let's start with H still to H is going to be cool. Tol phone 17 times 10 to the negative six Initial H 0.28 meters Times are changing temperature 28.5 degrees Celsius So dealt H is equal to 1.36 times 10 native four meters Now for hell tell tales equal to again. Same of and linear expansion coefficient. 17 times in negative, six times 1.34 meters times 28.5 degrees Celsius so are changed in L. A is equal to 6.49 times 10th negative four meters. Now if we want to find the magnitude of how much the elbow moves displacement we need to put in. Both are dealt age and our delta l square them add them together and put him under a square root. Find magnitude. So when we plug in our numbers built H 1.36 times to negative four squared plus till tol 6.49 times 10 native four squared should find the magnitude of displacement I'm just gonna leave this de for displacement is equal to 6.63 times 10 the negative for meters Now for the direction we could just find an angle Fada and we can find a direction through the inverse tangent It's kind of confusing that hell is in the y direction and h is an ex direction This example because you would figure ages height, which is why direction, But we kind of need toe change. These are inverse tangent three flipped him around, right? So fan is the inverse tangent of still toe l over and Delta H sequel to the interest tangent. Uh uh, 6.49 times 10 negative. Four over 1.36 times Teoh to the negative for. And you should find that are position, or I guess direction of this placement in this case is going to be equal to 78.2 degrees.

We have Tada equals are handed off Delta y over Delta X equals our Italian off. Does the wife is? Why not Alfa Delta t over X, not Alpha Delta T equals Dan Inverse. Why not over X? Not so we find theta it Waas arc tangent off 1 34/28. It was 78.2 degree Andi Delta are It was squalid off Tell the X Square. Plus, does the wife's grab equals X? Not Alfa Delta t squared. Plus, why not? Alfa Delta T squared square dude equals X, not square. Thus why not square square dude times? Alfa del GTI, on putting the numbers in this land would be 0.663 millimeter. So the extension is this must to the right on this much below horizontal.

So we have two pipes and an elbow joint. Call Vertical length L one and the horizontal length L two. Here's our all right. Now we know that l one equals 134 centimeters. L two equals 28.0 centimeters and the initial temperature is going to be 18.0 degrees Celsius in the final. Temperature is 46.5 degrees Celsius and finally we know it's copper. That's copper pipes and it thermal expansion coefficient is 17 times 10 to the negative six degree Celsius embers. So what's gonna happen when you explain when they go from the initial temperature? The final temperature is you can draw with the daughter. Lines are initial position. These two links air going, Teoh increase and move this elbow. So this daughter wine iso one. This horizontal dotted line is L two. These new ones are L one plus Delta L one and L two plus Delta L two and this length year. L What well is what we're trying to find. So first we need to find out what delta l two and L two l one r and we can just use the formula don't l equals Alfa Times L A Times Delta t so dealt is gonna be the same for breath. Um, so we'll calculate that right away t final minus t initial, which is 46 46.5 degrees Celsius minus 18.0 degrees Celsius. And that is 28 0.5 degrees Celsius. Snubbing calculate dealt. One is going to be 17 times 10 to the negative six degrees Celsius in verse times 134 centimeters times 28.5 degrees Celsius. We get 0.649 centimeters. Delta L two is going to be same feeling formula, but with a different initial else This one's gonna be 28 when zero centimeters times 28 25 degree Celsius to give us 0.136 centimeters through the changing links. But what we need to find is l and the angle in which it changes. So if you look down here free, zoom in right here. What will actually find is that IHS the elbow starts here and finishes here. This vertical change is delta one and the horizontal change is just delta L two and so if we complete this right triangle and we can call this data and try and find this data and this is a little well, So now we can issues Pythagorean theorem to find El and say that So we use President Diem is l Squared People's Delta L one square was Delta No. Two squared or l is just the square root of Delta No. One squared was Delta now two squared, playing in numbers Oh, 0.649 centimeters squared plus 0.136 centimeters squared. It's where it all of that and what we get is zero points 0663 centimeters or 0.66 three millimeters. Pretty small change. So that's L. And then for data, we can see that tanginess data is equal to Delta L one over Delta L two and we could take the are tan of both sides of salt for data. So the data equals Arc tanne. I was Delta l one are 10 of those 01 ever delta l two and plugging in numbers. We get Arc tanne zero point zero 649 centimeters abided by 0.136 centimeters. To get 78.2 degrees is data


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