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Velocity of the rock a5 function d. Express the instantaneous ~5.74 of time: (t ) = What is the instantaneous velocity of the rock at seconds? 7,5 6 m | 5 When does...

Question

Velocity of the rock a5 function d. Express the instantaneous ~5.74 of time: (t ) = What is the instantaneous velocity of the rock at seconds? 7,5 6 m | 5 When does the rock hit the ground? What iS the instantaneous velocity of the rock when it hits the ground? h'( 8.64 ) = /5 3.72 (6,Cl) Express the acceleration of the rock a5 a function of time: When is the instantaneous velocity of the rock equal to zero? 4'C What is the velocity of the rock when its height is 25 meters on the way u

velocity of the rock a5 function d. Express the instantaneous ~5.74 of time: (t ) = What is the instantaneous velocity of the rock at seconds? 7,5 6 m | 5 When does the rock hit the ground? What iS the instantaneous velocity of the rock when it hits the ground? h'( 8.64 ) = /5 3.72 (6,Cl) Express the acceleration of the rock a5 a function of time: When is the instantaneous velocity of the rock equal to zero? 4'C What is the velocity of the rock when its height is 25 meters on the way up? (2.35 ) = 6.17 mls What is the velocity of the rock when its height is 25 meters on the way down? (5.71) In /SC On the real test, the function _ derivative that = can be found may be replaced by any function usine the having but the questions ~kwill be differentiation rules in chapter essentially the same except Section 3.9 Related for the Rates ston:. spotlight on the ground shines on a wall Ifa man who is 2 meters tall walks 12meters - away from the from wIlat a speed of 1.5 meters per spotlight toward - second, how the his shadow on the building fast is = decreasing the when he is 4 ' length of thewall? meters from Or the real test; this problem may be replaced by the exercises in section 3.9 problems 17-29 any similar texk Lstarting problem page 249 like of Triola'$



Answers

The formula for the height of a projectile is $$s(t)=-16 t^{2}+v_{0} t+s_{0}$$ where $t$ is time in seconds, $s_{0}$ is the initial height in feet, $v_{0}$ is the initial velocity in feet per second, and $s(t)$ is in feet. Use this formula to solve.
A rock is launched upward from ground level with an initial velocity of 90 feet per second. Let $t$ represent the amount of time elapsed after it is launched. (a) Explain why $t$ cannot be a negative number in this situation. (b) Explain why $s_{0}=0$ in this problem. (c) Give the function $s$ that describes the height of the rock as a function of $t$ (d) How high will the rock be 1.5 seconds after it is launched? (e) What is the maximum height attained by the rock? After how many seconds will this happen? Determine the answer analytically and graphically. (f) After how many seconds will the rock hit the ground? Determine the answer graphically.

In this situation we have to find the velocity of object as the function of time here or a here equal to negative 32 v of t equal to integrate negative 32 duty. Then we get we of t equal to negative 32 t. Let's see now we put here to equal to zoo. Then we get zero equal to negative 32 two times zero plus c. So here you can see that's equal to zero. So three of t equal to negative 32 time t so you can see that velocity V is the integral of a at equal to zero. The entitled Real City is zero to which is our finance or for part a. Now we sold part use. The result is part A to find the position function. So the position function as is the integral of we eight equal to zoo. The entitled heights is 400 ft. So as of t equal to Uh huh intrigued Negative 32 Time T DT. So you get that's of t equal to negative 16 t square. Let's see no s of T equal to 400 ft and equal to zero So we put here and we get 400 equal to zero. See? So here, sequel to 400. So now we put here as of t equal to negative 16. The square plus 400. So it is a final answer for Part B. No, we sold part C. Yeah. Here. Equal to square root 32 upon. Okay. Which is even in discussion and you equal to be. So now do we upon 32 Negative. Okay, three square equal to negative. Intriguing. Didi. So you can write like that one upon k in trouble DV upon 32 upon key Negative, we Squire equal to negative intrigue duty here. We know that in trouble. Do you, Paul? He square negative. We square equal to So here is you square equal to you. Want to pawn to a the natural law? More release e Plus, you found a 92 Hugh plus c. So we apply here and we get one upon two times g time. 32 phone. He's quite 32 upon natural square root 32 upon k plus we Oklahoma, the square root 32 upon. Okay, negative. B here is model A say equal to negativity. No, We simplify this and we get natural. Oh, the more you lose. Square root. 32 upon k. Close we upon square root. 32 upon. Okay, negative. Three. Equal to negative. Two time Kay Square Root. 32 upon Katie. No square. Italy too. Bon ki plus three Capone. Okay. Square 32 Phone. Okay. Negative. B. He called to into the power. Negative two is square root. 32. Katie. Now you can write like that. One of one into the power to Times Square root. 32. Katie, no, We do cross, multiply and we get square root 22 upon. Okay, It's time into the power. Two times. 32 k p. Place we time. It has the power to time squared 32 upon. So he Katie, 32 time Katie equal to square root. 32 upon k. K. No, you do we? No, You can see that. Here we which is this year and here. Break it it the power to square dude. 32 Katie plus one equal to 1932. So he square 32 upon k. It took the power two times the square root 32 Katie. Negative one. So here you can see that we could do negative square root 32 upon. Okay, time into the power to Times Square Root. 32 Katie place. So the negative one upon into the power to square root. 32 Katie. Positive one. So you can see that here. We of t equal to negative the square root. 32 upon key time. Hyperbolic 10. Okay. Square road 32. Katie. Mhm. So it is a fine answer for part C? No, he sold for D. It is easier to evaluate if we look at the e form in the part with a has become one living negative square 32 upon K, which is called terminal velocity. So here limit d tends to infinity now you too. Square root. 32 upon k into the power two times 32 key. The negative one a phone he to the power two times 32 g d for the 21 So you can see that Here we get negative square root 32 upon k here. We know that we put here to equal to in free day. So we get here one so we can say it is a final sell for Part D? No, we saw party here. Sequel to 400 since and equal to zoo displacement is 400. Like we did the displacement equation without air resistance. So you can see that here s of t equal to negative square root 32 upon k and kill proper Bali. Mm. Then oh, 32 Katie d t. So here. We know that. Now. You, too. Square root. 32 upon k in trouble. Hyperbolic sine of the square root. 32 Katie upon hyperbolic course, the square root 32 Katie duty. So, no, we simplify this. And negative square root 32 upon. Okay, Now we get the integration of hyperbolic sine of square three to Katie, upon hyperbolic calls of the square root three to Katy. So we get one upon square root. 32 Katie Natural hyperbolic Course. Oh, the square root. 32. Katie, we'll see. So here we know that's equal to 400. So we put here and we simplify this and we get negative one upon k natural hyperbolic cause of the square root 30 to Katie, plus 400. So it is. Our financing, for part is no, It's all part. And you can see that in this graph neglecting resistance blew the object trees ground to equal to five with a therapist, and the object reached the ground eight equal to here. Approximate value is 8.3. So additional time is about D additional. Right? Okay. Is about 33.3 2nd. Mm. In the part F. Yeah. You in Greece, K in the time to reach the ground will increase with Okay. Cool, too. 0.2 mhm. 02 The growth so confirmed it. So it is all files.

So in this question, We are told that a stone is thrown upward to the speed of 20 m/s and it comes down and it's caught five m above where it was thrown. And we want to find the velocity at that point where was caught. And we're going to use the formula V. F. The final velocity squared is equal to V. I squared. Initial velocity squared Plus two times, actually transmits gravity times the red skull distance wire. And instead of trying to Kathleen all of this distance and trying to calculate the height and trying to find the velocity here, Why don't you just consider the motion of the remaining five m? It felt that you should have fallen if it wasn't caught. And so there's going to be the initial velocity V. I. That's unknown. And the final velocity, the velocity at this point And philosophy, finance lost at this point is going to be 20 m/s because when the stone is thrown up, decelerates at 9.8 m per second squared and which is the point, the top maximum height at zero m a second and then accelerates down back at the same rate to to end up reaching the same velocity. And so Our Y is going to be five m And G is 9.81 meters per second squared. And so substituting, we have 20 squared you called to V I squared Plus two times 9.8, one times y she's five and B I squared V I will be equal to a squared uh 20 squared minus two rocket nine point each one times five. And so uh initial velocity at that point is going to be 17. It says best second. Yeah. And so I've lost at this point is 17 m/s. Mhm. And now for the time we're going to use the formula, it records with me f minus B by over tea and making T. The subjects we have T. Is equal to V. F minus B. I over A. And this is this will be G. Because we're extending acceleration due to gravity And this is going to be the final velocity 20 minus in travelocity. When I went to use the approximate that value, we're going to use the original value which is 17 point 37 5-6 over Jeanne. My point it one. Yeah. Actually me making a mistake here. The time should be the time it takes a look at what time to the time of flight. For the time it takes to reach here and the time it takes to calm down to this point. So we can find the time it takes to go from 22 0 m per second squared per second And accelerating at 9.81. And v. F. in this case is going to be zero minus The initial velocity, which is 20 of a team. We're taking up as *** direction, so -9.81. When it's going up, you got c minus nine point it's one. And And this is 2.04 seconds, you have the fresh box and then we need to find this time it takes you get down here. So finally the time it takes to go from philosophy of Ciro, so zero To have lost year 17 four, it was second. The full value is going to be 17.37 5-6-0 of our Over the exploration of 9.81. But taking down this positive direction And 9.81. Yeah. And the results for this is 1.7 seven second. And adding these two together. We get our total time for flights. Yeah. Cook the side So time T. Is equal to 2.04 Plus 1.7 seven is the culture 3.81 seconds. And we can approximate that to 3.8 because we're dealing with two decimal places and all our figures 3.8 seconds. So this time and the mary sits by the side here at So the we're dealing with two significant figures. Excuse me, not to the small places. So the velocity at that point Ghostee at five m Is it be called to 17? That's the second and at the time 3.8 seconds to the time of flight.

Mhm. Okay. What we're gonna do is we're going to walk through a kind of like a word problem. Um We're starting with an object that is projected upward from ground with an initial velocity of 500 ft per second. So this is a vertical motion problem. And the first thing we want to do is we're actually going to be neglecting air resistance. And what we want to do is to determine the velocity equation. And so we know that um the velocity equation generically is our velocity is equal to our initial velocity plus the acceleration times time. Okay, so for our purposes um The initial velocity we're told was 500 ft/s. And of course acceleration because it's vertical motion acceleration is due to gravity, so that is negative 32 ft per second squared. Um So it's going to be negative 32 times T Okay, so there is our velocity function with respect to time and now what we're gonna do is we're gonna actually graph um this velocity function and of course um this velocity function is linear. So this is an equation for a line. But we're gonna go ahead and grab him using a graphing calculator. So I'm gonna switch over to my graphing calculator. Dez knows um is it online, free online graphing calculator. Um And we notice already have it in Dismas. Um And so I'm actually going to um Zoom out. Um and so you notice of course the Y intercept is right up here at 500 and it's a decrease in line. So this is what that decreasing line looks like right here. And now I'm gonna go ahead and quickly sketch my graph. And so if this is T. And this is my velocity Moded at 500. And then somewhere here around 15, a little over 15 is where it crosses. So we're just gonna go ahead and do that. Um Okay, so that is the velocity function. And so now the next thing we want to do is we want to go ahead and use our velocity function to find the position function. And so remember, the position function is equal to the integral of that velocity function. And we're integrating of course, with respect to time. Okay, um and so now we're just going to integrate this. Um and so my position function Is equal to 500 t -30. Open note is not going to be 32, it's gonna be 16 T squared plus some constant of integration. Okay, so now we're going to find that constant of integration. Right? And so what we know is The object is projected upward from ground. So at time zero at a time Zero, my position is on the ground which is at the zero level. And so that is going to be equal to 500 Time 0 -16 times zero Squared. So that is telling me that constant is zero. So really my position function is equal to and I dont know why thats freezing up on me is 500 t minus 16 T squared. And there is my position function. And so now we want to know when at what time do we reach maximum height and what is our maximum height? And so remember, time at max height is when my velocity equals zero. So we're going to set our velocity equal to zero and we're going to sell for that time. So that tells me t is going to be equal to about 15 .625. And so now I want to know what my maximum height is. So I am, my max height is equal to my position function Evaluated at that time of 15.6- five. And of course this is in seconds. And so this will be 500 times 15 0.6 Too far to five. I don't know why that's doing yet -16 Times 15.6-5 squared. And so that will actually be equal. If you put that in your calculator, that is actually going to be equal to 3000 906 point to five ft. So there is my max height. Okay. And now what we want to do is um now we're gonna look and this is if I neglect air resistance. Right? So now what we want to do is um if we're told if air resistance is proportional rips to the square of velocity, okay? Um and it's going to be given by this equation that the derivative of the velocity is equal to negative 32 times. And that should be on the outside. So negative. And in parentheses we have 32 plus K B squared. Okay. Um and we want to use, we want to actually solve this integral equation. So we have one over 32 plus K V squared. D V is equal to negative the integral of D. T. Okay. Um this denominator almost almost looks like an arc tangent. That the problem is this K. That is attached to the U. Squared. Or in our case R V squared. So if I factor out a K. So I'm a factor out. I want over care. I get the integral of one over 32 over K plus B squared. Now that definitely looks definitely looks like an arc tangent now. Um So this integral of DT Okay. Um and so that definitely now looks like one over a squared plus U squared, which is an arc tanne. Right? So um that integral um is a one over Okay. Times and of course the integral um of one over a squared plus U squared is um one over a times arc tangent of you over A. And so that will actually equal negative T. Plus. See okay, so now what we want to do is kind of rewrite this because we want to solve for velocity, we want to have this in terms of velocity. So we're gonna kind of clean this up a little bit um and get velocity by itself. And so we and I do that. Um We end up with this actually becomes one over The Square Root of 32. Okay, and then this will be arc, that's two arc tangent of and then this is the square root of K. Over 32 times fee. Um Equal to negative T. Plus E. Okay. Um and so now we're gonna multiply both sides by that square with just 32. And we're going to solve for T. I'm 40, I'm sorry. And so when we do that we get the is equal to the square root of 32 over K. Times tangent. Because remember we've got to do the inverse of art tangent and C minus. And then we have the square root of 32. Okay. Times T. Okay. Um And now we need to find out what C. Is right. We want to find out what C. Is. And so we want to go ahead. And what we do know is when t when T equals zero, then the Is equal to 500. So if I use that concept and save 500 is equal to. And then We put in here the 32 over K times tangent. So that's just going to be equal to tangent of. See because T. is zero and so we get C. Is actually going to be equal to Yeah. Arc arc tangent. I don't know why this is messing up. Arc tangent of 500 times the square root of K. Over 32. Okay, so now if I put that all back in here we have the velocity as a function of time is equal to Yeah, Is equal to. And then we have this big old long thing, we've got the square root of 32 over K. Times tangent of C. Now is arc. Mhm. Are tangent of 500. Okay. over 32 and then that is going to be minus The Square Root of 32. Okay, times there we have it. That is a big old long crazy mess. Um And then what we want to do is we actually want to um use a graphing calculator. And we're actually gonna graph, we're actually gonna graph mhm. V. F. T. When K. is equal to .001. So we're actually gonna put K. In there as 0.001. And we're actually going to graph VF. T. And from that we actually want to approximate the time zero in which The height this maximum. Okay so wherever there's AK. We're gonna insert .001 and we're going to graph it. So I'm gonna switch back to um my online graphing utility. I'm gonna turn off that linear function. And you notice I already have um this function here right here. And I put in um Where there was, okay, I put in .001. And so I end up getting um I end up getting let's see um 32,000 square to 32,000 times tangent of art can um of 500 times 5000.3125 minus that 0.32 times the X. Value for the T. Value. And here is here is that graph and I'm going to be focused on this section right here. And so you notice that where my height is going to be maximum is the area under the curve right? Um And it's going to have to be a positive area under the curve. And so if I just kind of look at this graph, if I keep blowing up this is kind of like a tangent function. So I know where that area is going to be. Maximum is in this little triangle spot right here from zero to some X. Value. Or in this case t. So that is telling me that maximum height is going to occur when t. Is equal to 6.8 six. So if I switch back over here whips where'd it go? So if I switch back over to my right board. Okay. So um so when K equals .001 um max max height will occur at T equal to 6.86 seconds. Okay. Um and now what we want to do um is used an integration capability to actually find that max height. So we're actually going to integrate from 0 to 6.86 of this function right here. Okay. Um and of course um with K Equal 2.001. And so we're going to actually integrate and we're not gonna do it by hand. Of course that would be totally crazy. So we're gonna do um integrate of I think it was mm 32,000 then it was tangent of are tangent Of and I think it was 500 times the square root of point 1234 and I think it was 3125. And this was all in the arc tangent minus the square root of .03 two T. T. T. Okay, so we're actually going to use the graphing the capability um integral capability to integrate. Um I already have this in here so I'm actually I'm going to I'm just actually changed this and I'm going to do in control from zero. That's the neat thing about gizmos 8 6. And then of course I've got to add in my um the X. Here. Okay and that is saying my max height is 1088, so 1,088 ft. So based off of the integral, so this is about 10 88 feet. And so why is there such a big difference between what we found in part a versus what we did here? And the big difference is in part a or the previous owner where we neglected air resistance. Okay, so in in the previous part, we had no air resistance. And remember, air resistance is going to impede the vertical motion of your object. And so since this latter part, we actually inserted air resistance, we would not expect the object to travel as high because of the resistance of air resistance, the resistance of the air friction.

Today is what is called vertical motion. Um We're going to start with an object that is projected upward at 500 ft per second from ground level. And so the first thing we want to do is we want to neglect air resistance because that um as kind of an additional component um to the problem. So for right now we're gonna neglect air resistance. And the first thing we want to do is to write the velocity equation. And so we know that the velocity equation with respect to time is given by the initial velocity plus the acceleration times time. And for this problem, yeah, our initial velocity is 500 feet per second. Um It is vertical motion. So the acceleration is due to gravity which is negative 32 ft per second squared times time. So there is our velocity equation. Um And now what we want to do is actually to graph this velocity equation and you notice that it is um linear. So we could quickly do it by hand, but we're going to go ahead and do it on a graphing calculator first just to get a very good picture of it. So I'm gonna switch to gizmos, which is an online graphing calculator and already have it inserted into my graphing calculator. And so this is a linear function. So we know it crosses the y axis at 500 and um it almost looks like it crosses the X axis a little over 15. So let's go ahead and just go back to our white board and we're just getting quickly graphic. And so this is time if this is 15 and this is my velocity at 500. We know what kind of crosses a little bit here and here whips. And so there, we sort of have our linear function. Um And then the second thing we want to do is we want to use this velocity equation to find the position function. And so we know that the position function is given by the integral of our velocity function with respect to time. So this is gonna be the integral of 500 minus 32 T. And we're integrating with respect to time. Yeah, so my position function is going to be 500 t minus 16 T squared plus some constant of integration. And so we want to find what that constant of integration is. And so we do know that at time zero then our position function evaluated at zero is also zero. So if I insert that in here I get 500 time, zero minus 16 times zero squared plus C. Which tells me that C. Is equal to zero. So my position function is 500 t minus 16 T squared and that is it. And I want to use this to find my maximum hype. And so I know my max height is going to occur when my velocity is actually equal to zero. Um And so I'm going to use that to find the time because in order to find my position, I need time. So I'm gonna set my closet the function equal to zero and solve for time. And when I do that I get 15 point 6 to 5 seconds. And so when I find my position at 15.625 seconds, I'm going to find my maximum height and I believe that comes out to about 3906 point 25 feet. So there is my maximum height. Okay. Um and now this the second big thing we're doing because remember all of this was based off of neglecting air resistance. And so now um we are told that air resistance is proportional to um velocity squared or the square of the velocity. Um and we're given this equation that DV over DT is actually equal to negative 30 to minus K V squared. And were also given the equation where we're actually want to solve, we want to solve this equation. So we have the integral of one over 30 to minus kv square D V is equal to negative the integral of DT And so basically I just decide dividing both sides by 32. Um this is minus K V squared. Um and then this is some kind of looking like um the integral of 1/32 miles K V squared is almost looking like a I think it's going to be a plus um is looking like an arc tangent. What's throwing yourself? Is this K. So if I factor at a K, we get 32 over K plus V squared tv. Okay, so a squared is 32 over K. And then of course this would be V squared is actually U squared. So this now is an arc tangent and so we know um we have one over K times one over A. Which is going to be one over the square root of 32 over K times arc tangent of V over a. Which is that 32 over K. And that is going to be equal to negative T. Plus. Some constant of integration. Okay. And so now what we want to do is actually saw this for the so we want to go ahead and solve this for V. Um and we can simplify this. This is going to be one over the square root of 32 K times are tangent of the square root of k. Over 32 times fee minus is equal to negative T plus E. Okay. Um and then when I saw this for V I get V is equal to the square root of 32 over Okay, 32 over K. Um Yeah, times tangent because I have to do the inverse of art tangent to get the out of the um hit be out of arc tangent. So this is going to be tangent um of C minus the square root of 32. Okay, times T. And so there is V right here. Um And so I need to find out what C. Is and so in order to find, see I know that when T equals to zero, my velocity is 500. So I'm going to use that concept. So am I said V equal to 500 and then I get 32 over K times tangent of C. Because T zero. And then solving for C, we get C is equal to arc tanne of 500 times the square root of Okay, over 32. Okay, so now putting that back into my velocity equation, I get the velocity with respect to time is equal to the square root of 32 over K times tangent of an arc tanne as 500 times the square root of K. Over 32 minus the square root of 32 K times T. Okay. Um And so this is actually our velocity equation. Now what we want to do is we want to let K equal 0.0 01 I believe. So when I do that we get velocity with respect to time is equal to 32,000. Right tangent of arc tangent. Um This brought in here actually becomes um I believe um let's see um 500 times the square root of point 0000 And I think this becomes a 3125 in here. If I just minus the square root. And I think I have yeah, the square root of point 00 point 032 times. T Okay, now the next thing we want to do is we actually want to grab this. We want to actually grab this to find the time of max height. Okay, so let's switch back to our dez mose and I've already got it in here. So I'm gonna turn off my first one and turn on this one. So this is my graph. Um And of course it's a tangent grass. So it's going to look like a tangent graph that has shifted. And so our max height is going to be occurring. Where are we have maximum area? Right? So if this is my velocity graph, then when I um find the area underneath the curve between the X. Axis and the curve that will be um my position. And so if I calculate that point right here, which is 6.86 that is going to be the time for that max. Hi. And so if I come back here, I'm going to have t is equal to 6.86 seconds. Okay. So now what we want to do is actually find that max hi. And that's going to be equal to the integral from zero to their 6.86 seconds of this velocity equation right here. So of V. A. T. DT. So and we're not gonna do that by hand by any means. So what we're gonna do is actually go back to our graphing calculator and we're going to use the integration feature. Um And so what I'm gonna do is go back here and instead of using this as a graph, the neat thing about dez moses I can just go ahead and delete that and then just start typing in the word integrate. And I'm going to go from zero two six point 86 And of course it's not giving me anything because I haven't told it that I am integrating with respect to X. And so now my integration is 10 1000 and 88 ft. So is going to be 10 88 ft approximately. So why is there such a big difference between um my initial maximum height and this maximum height? And it has to do with air resistance. Remember air resistance is going to resist um the object going upwards. Um So it's going to diminish how far it can go upwards


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Kdtogen bjnoing#tongorsImmoeu jorcessubutdncer Okeitix molal MasrLonoot Eneaslon (CicermeleculezDibala DiEale 'ore atangoIncroising moleculnconond {orcecnyooer
Kdtogen bjnoing #tongors Immoeu jorces subutdncer Okeitix molal Masr Lonoot Eneaslon (Cicer meleculez Dibala DiEale 'ore atango Incroising moleculnc onond {orcec nyooer...
1 answers
A neutral pion $\pi^{0}$ (rest energy $=135.0$ MeV ) produced in a high-energy particle experiment moves at a speed of $0.780 \mathrm{c}$. After a very short time, it decays into two $\gamma$ -ray photons. One of the $\gamma$ -ray photons has an energy of $192 \mathrm{MeV}$. What is the energy (in $\mathrm{MeV}$ ) of the second $\gamma$ -ray photon? Take relativistic effects into account.
A neutral pion $\pi^{0}$ (rest energy $=135.0$ MeV ) produced in a high-energy particle experiment moves at a speed of $0.780 \mathrm{c}$. After a very short time, it decays into two $\gamma$ -ray photons. One of the $\gamma$ -ray photons has an energy of $192 \mathrm{MeV}$. What is the energy (in $...
5 answers
What are the targets of SIRT1?
What are the targets of SIRT1?...
1 answers
The following is the distribution of instructions within programType of instruction % of program Arithmetic CBZ/CBNZLDUR STURThe following is the accuracy of the following % branch-precliction methodsAlways-taken Always-Ho-taken Dyamic prediction IS% 52k 837AsSume (at mispredliet will cauSC stall of % clock eyeles_ What is the approximate eXtra CPI due t0 mispreclicted branches with each of thie branch predliction methods? AsSue tat there are Ho dlata lazards_
The following is the distribution of instructions within program Type of instruction % of program Arithmetic CBZ/CBNZ LDUR STUR The following is the accuracy of the following % branch-precliction methods Always-taken Always-Ho-taken Dyamic prediction IS% 52k 837 AsSume (at mispredliet will cauSC sta...
5 answers
1f 405 X (023 molocules Suhslork ( havc mOSf cl 86.29' hha IS Ihe malG t mSs cl tha Subslantc Iccv
1f 405 X (023 molocules Suhslork ( havc mOSf cl 86.29' hha IS Ihe malG t mSs cl tha Subslantc Iccv...
5 answers
X= Identify Find the (Use all the equilibria li to of the separate differential sjamsue equation as Xp needed:) 10 3x + X
X= Identify Find the (Use all the equilibria li to of the separate differential sjamsue equation as Xp needed:) 10 3x + X...
5 answers
Apoln that contains an acctal functional group (by clicking in the appropriate box with your cursor} Selert the compoundCH;OHOHOHNextpriilout
Apoln that contains an acctal functional group (by clicking in the appropriate box with your cursor} Selert the compound CH; OH OH OH Next priilout...
5 answers
6. Find the derivativesa.y = tan 3x - cot 3xb.y = csc" (Zx)sinx cy =
6. Find the derivatives a.y = tan 3x - cot 3x b.y = csc" (Zx) sinx cy =...

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