Question
Practice Problem Submissiong are not pemmanently recordedThis problem practice problem , attempts are graded May practicc This problem has seven parts (submlt buttons); cnecken separately:problem repeatedly by selectingProblem Variation,Considermolc ~ampic 4mMntie pressureQa; undcrgocschanac3tatl-Dres:uIc20 bar; ASSuMIC the gas bchaves ideally;prccessleypansloncompressom The viork associated Hith this pccess(posiye neoaiiyezero :Submil AnawarCelculate the wot k the process Submit Answver riescau
Practice Problem Submissiong are not pemmanently recorded This problem practice problem , attempts are graded May practicc This problem has seven parts (submlt buttons); cnecken separately: problem repeatedly by selecting Problem Variation, Consider molc ~ampic 4m Mntie pressure Qa; undcrgocs chanac 3tatl- Dres:uIc 20 bar; ASSuMIC the gas bchaves ideally; prccess leypanslon compressom The viork associated Hith this pccess (posiye neoaiiye zero : Submil Anawar Celculate the wot k the process Submit Answver ries cauied ut inevensib aqainsi constant exernai Wicssure) ano isothenally against the = 20 bar, the final DuM Suppose expanslon occums Two steps In this case the magnitude viork will L (greater; lesser equal than to the magnitude work for one-step expansion betwveen same Initial and final states. Submil Answbr Trles 073
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Each of these problems consists of Concept Questions followed by a related quantitative Problem. The Concept Questions involve little or no mathematics. They focus on the concepts with which the problems deal. Recognizing the concepts is the essential initial step in any problem-solving technique. Concept Questions In a performance test two cars take the same time to accelerate from rest up to the same speed. Car A has a mass of $1400 \mathrm{~kg},$ and car $\mathrm{B}$ has a mass of $1900 \mathrm{~kg} .$ During the test, which car (a) has the greater change in momentum, (b) experiences the greater impulse, and (c) is acted upon by the greater net average force? In each case, give your reasoning. Problem Each car takes 9.0 s to accelerate from rest to $27 \mathrm{~m} / \mathrm{s}$. Find the net average force that acts on each car during the test. Verify that your answers are consistent with your answers to the Concept Questions.
In this problem. We are going to calculate the increase in temperature of photo and the system reaches two the thermal equivalent. So we have to calculate here T minus two w That's right. The equation. To calculate this, uh, increasing temperature or fortune as a capital M is cf into few minus t f plus m w C w into the change in temperature and water that is equal to t minus T w This is equals to zero l s m f is the mass of the a total mess of the spent fuel assemblies. CF is the especially the capacity of the assemblies and spent assemblies fuel assemblies and this is the change in temperature. Similarly here this M w is the mass of water C W is the specific heat capacity of the water and this one is the change in temperature of the water. So from here we can wipe the equation for the equivalent temperature as a T Z equals two capital M f uh c f gs plus, uh, M W C W. And here we have 40 w divided by a small M f smaller, smaller F C f plus uh, M W C W. This will be a record number one. Let's put the values into this equation. Number one Jay Z equals two square books. Clothes open. Here we have the mess of total mess of the assemblies that is equal to 184. This is the number of families, and the mess of one assemblies equals to 320 two kg into C F, which is equals to 289. Uh, 40 84. Here we have 284 Joel Parkinson for Calvin into TF, which is equal to 650 eight plus 270 three into Calvin, plus the next term. Second, down here we have the mass of the water, which is equals to 101,720. Multiply with Terrence Bar, three kg into 4184. Joel parquet cheaper Calvin into the a temperature of water which is equals to 15.0 plus 273 into Calvin. We have to close the square rips and we have to divide this at this town. Uh, with 200 48 into 322 kg. So this is the total mass of the spent fuel assemblies, uh into it's a specially capacity that is equal to 284 jobs per kg for Calvin Plus now the mass of the water which is equal to 1700 and 20 multi collaborator is about three kg into 4100 84 Joel Parkinson per kelvin square books, clothes. So from here we can get the value for the 30 as it is equal to 200 90 Calvin. Now, this can be written in terms of the degree Celsius as this t Z equals two, uh 290 minus 2, 73 into degree Celsius. So from here, we can wind this temperature as it is equals to 17 degrees Celsius. So this is the equivalent temperature. Now we can wait this t minus t w which is the increasing the temperature of water. This would be equals to 17 degrees Celsius minus 15 degrees Celsius. So we have two degrees Celsius. We can see that That is an increase of 2.0 organizations Temperature in the water. Thank you
And this question, we have a rod that is initially 3.682 m And expansive 3.704 m. And we're looking for the increase in its length, so the increase in its lengthy, the final length, modesty, initial length. So we have 3.704m -3.682 m, Which is equal to 0.022 m. And we can convert this length and two where the prefix milli lowercase and is equal to 10 to the negative third. So one m, he hold to a 1000 or one millimeter is equal to 10 to the negative third meters. So in order to convert meters two millimeters, we'll start by writing zero port 0 to 2 m. And we need to multiply by a fraction that has meters in the denominator. So then the unit will cancel. So we can write one m in the denominator and we know that's equal to 1000 millimeters. So right that in the numerator and you can see here that meters will cancel out, leaving us with just millimeters. So we have 0.22 times 1000, which is equal to 22 millimeters, so the increase in the length is 22 millimeters.
Way have initial momentum plus f t. It was final mo mentum. Final momentum is M V. Final on initial is M v I. So we find the final. It was three initial. Thus f t over em putting the numbers in This will come out as 20 i seven j meeting for a second acceleration is Delta V over Delta T. It was four i middle per second squared.
In this problem, we are going to calculate the minimum amount of water that can bring the system at any temperature below 40 degrees Celsius. So in this case, we have to calculate mass of water. Let's write the equation which show the balance in in the heat floor that can be written as M s, which is the mass of the scoot CS Delta tears his equals two M w c W delta T w Here. This m s is the mess of the stock. Our skilled CS is the specific heat capacity of that skill, which is made up of uh l monium Delta T s is the change in temperature in the skillet And this, uh, m w is a mass of water. C W is the specific capacity of the water, and in this delta T W is the change in temperature in water. All this equation can be written for MW as M. W Z equals two uh, M s CS Delta T s you wanted by here We can white c w delta T w. This will be our equation number one. Let's set the values into this question, so M w is equals two into 2.33 kg into CS, which is equals to 900. Your package per kelvin into the change in temperature, which is equals to 286 minus, uh, 14 to Calvin. Divided by here we have C W, which is equals to 404,180 for General Perkins Super Calvin into the Change Interpreter, which is equals to 40 minus 25 into college. So from here, we can get the value for this. M W S M W is equals two 8.22 kg. So this is the required mess of the water that is needed to bring the system including temperature. Thank you.