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[-/1 Points]DETAILSSERPSE1O 7.5.OP.0z0.CTX:MY NOTESPRACTICE ANOTHERYov Are working Tacony With nuraer Iaiqe mechanical ulachines Mlachine involvec plunuer Mdys 00 k...

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[-/1 Points]DETAILSSERPSE1O 7.5.OP.0z0.CTX:MY NOTESPRACTICE ANOTHERYov Are working Tacony With nuraer Iaiqe mechanical ulachines Mlachine involvec plunuer Mdys 00 kg that moves horizontally into complex array gears; Jctc sprngS Your supervlsor wants make 5ome adjustments Tensions ano othe parameters assoclated ulth the amav comoonents He clalms that hls adjustments wlll result the plunge experiencing constant force maanitude (Owaro array comp onents while the array components will push back with

[-/1 Points] DETAILS SERPSE1O 7.5.OP.0z0.CTX: MY NOTES PRACTICE ANOTHER Yov Are working Tacony With nuraer Iaiqe mechanical ulachines Mlachine involvec plunuer Mdys 00 kg that moves horizontally into complex array gears; Jctc sprngS Your supervlsor wants make 5ome adjustments Tensions ano othe parameters assoclated ulth the amav comoonents He clalms that hls adjustments wlll result the plunge experiencing constant force maanitude (Owaro array comp onents while the array components will push back with position-vanyina rOrce magnitude mhere 55,0 ana Torces act at the same position On plunder lorce veclors maqnitudes exdcl opposite directicns The plunger begins moving inward at x ilhu speed 0l 2.50 m/s. By what distance (in m) does the plunge inward before comes rest? Need Help? MLJI= Submil Ansie



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(1I) A cyclist intends to cycle up a $9.50^{\circ}$ hill whose vertical
height is 125 $\mathrm{m}$ . The pedals turn in a circle of diameter
36.0 $\mathrm{cm} .$ Assuming the mass of bicycle plus person is 75.0 $\mathrm{kg}$ ,
(a) calculate how much work must be done against gravity.
(b) If each complete revolution of the pedals moves the bike 5.10 $\mathrm{m}$ along its path, calculate the average force that must be exerted on the pedals tangent to their circular path. Neglect work done by friction and other losses.

Hi friends. If must end Yeah. H is subject to as I just kept force my next baby again. Mm and we thought must be equal to minus B into velocity. So when the city you will get minus common to be here, you comma is be upon him. So the solution of this publicity will be we're not exponential of poverty. Yeah. Mhm. Back We're not to be scared judo. Not be but the equation of motion of. Mm. Yeah. Okay. For the driven motion X double dot plus Obama X dot is going to have not calls for Mikati here have not is have not. I am. Mm Yeah, putting the study strategic question. We will get Yeah. Mhm. Mhm. You could get minus pay or you guys squared because of mega T minus data. Okay. My name is gonna amigo. A sign up. We got t minus comma. He's got to have not course up negative. So this further can be simplified as All right. Yeah. Because of mega t. Okay. Yeah. A course of omega T Also up data minus. Come on. *** mm. Yeah. Yeah. Yeah. Into a course of mega T. Cost data plus. Yeah. Sign off. We got it sign of data. And to me guys squared let's go moe mingo signed off negative because of data minus course of negative. Sign up gamma. Yeah, Sorry, I have to correct. It's good to have not because of the negative. So this equation finally you can right cause of omega T into have not plus A When you guys square because of delta minus gamma omega sign of data, but less Yeah, sign of omega T. A. For me guys squared sign of delta. The less comma omega because of data is culture. Mhm. Do you know? So that too angular Brackets have to be separated equal to zero. Mhm. Yeah. Have to be separately equal to zero. So we will get sign of delta plus when you guys square sign up data into mega square plus gamma mega cause of data. So then of delta you will get mhm minus gamma upon America. That is Faith. You will get 10 in worse of my next come up on. Sorry come up on data. Yeah. And other amplitude you will get a school to have not divided by comma omega sign of victor minus for me guys squared or subject to and sign up data, you will get 10 of Taster divided by Multiple us 10 square of data. So people will be minus comma born omega one plus gamma square upon Amigas square or minus you can drive minus car back on router. Got my square. Let's make a square. Okay. Yeah. Now, because of data, you similarly because of data you can right bon upon route of one plus gamma square upon you guys square. So it is to be omega upon route of mega square plus go my square. So substituting the value of a comma cause of delta. You can write f not by him upon omega into burn upon omega is squared. Let's go on my square. Mm. The um the complete solution will be equal to X as a function of time will be a cause of when he got to you minus steel top plus c minus we not upon you comma explanation of minus common to T. And the velocity mhm can be written as extort as a function of table b minus omega. A sign off. We got my nest on top plus we're not exponential of my next committee. Okay. Okay. He because of delta is cool too. We're not going comma minus E. And omega a sine of data is called to minus B. Not Dividing equation to win one. You will get omega 10 of data. It's cool to minus. Maybe not. We not by gamma minus C the package. Okay. My name's comma. Oh okay. Yeah. So minus B. Not it's got to minus comma. We not gamma minus C. That is -7. Not gamma policy So C is equal to zero. So be not. You will get he come up because of data That is -1. If not. And when you go to walk off, when you guys square place gamma square and two and make up on route dog. Okay? You guys squared plus commas square so you will get a comma If not I am Okay guys square plus about my script and X key graph will be like this. Yeah. Mhm. That so thanks for watching.

Okay, so we know current. It can be able to be Manitou off the MF over resistance. Which is he gonna be? All of you are therefore have B. Is he who I are over LV and we know the my 90 force can be able to be out now. Can use IR over Albie substitute for beat. So have I r o l b test ir. As you can tell, I'll here to make ends well, and this will give us by any force. Is your high score r over V. It's not over. Those are Angela here. We'll have the current square is going after Mac tens be over are Therefore the currents should be able to square root. Uh, magnetic forces can speak over resistance. So the man any force is giving us one Newton speed is two meters per second and the resistance is given as up a omega. So now it's playing is where you spend two equation to determine the, uh, parents, which is equal to one Newton guys. Two meters per second over a point 00 omega. And this is you go to Joe going by. I m here. So now let's take a look and e r b Barbie was asking us the, uh rate at which energy deliver to the resistor, which is asking us the power delivered to the resisters. So we know power can be with my square are when occurrence 0.5 AM here, resistance a omega, not it's plugging these values bring to the equation to determine a power which is equal to zero point If I am here to a power to and times omega and this is equal to who wants so for Parsi in order to determine the mechanical power to deliver by the force half. Um, when did stand out such fortune here because of the mechanical power? Yeah, deliver by the force should be go to force tends to speak and we know the forces one Newton speeds two meters per second. So not as funny as writers back into the equation to determine the mechanical power. And this is equal to one Newton times two meter per second and this is equal to to us as well. Therefore, for the last question here. So based on the calculation, which is that we know that the, uh great, at which energy is delivered to the resistor should be going. So the mechanical power they were everybody force after. And these are the answers for this question.

So initially calculate the energy stored in the system. The N G U B is equal to 1/2 and in doctors l m the square of the current. And we had given these values so that Sahhaf and the inducted its on these coils. What's G. Henry Times I squared and the current 50 tens 10 to the three MPs. This is Old Square, and this gives us an energy stored off 6.25 times 10 to the power Tim Jules. Secondly, two adjacent tongues, a parallel wires and they both have current, which is moving in the same direction. And since the loops have such notch radius, a one meter section such as this can be the guarded as fairly straight. And hence one wire creates a field. Be easy. But, um, you know I over two pi r This field causes the force on the next wire and this force, if is equal toe I and so times be sign beatem where l is the length off the wire and hence we can get an expression for the force. Are you limp? That one wire inserts from the next. And this force Peanut Lin Yes, over l is I times you know it All right. Over two pi r sign 90 degrees Since they are no particular to each other and this gives us, you know it times I squid over to buy are And hence if we substitute our values we can find a force per unit Length to be you know it which is full pie times 10 to the minus seven in S I units, I squared. We know the current year's huffed e times 10 to the power three m piers squared over two pi r to buy new Rania's is zero point to five and this gives a force unit link both 2000 new tenants for every meeting.

For a turntable were given the initial angular velocity, which is Omega and is 1.8 revolutions per second, which I convert to radiance per second by multiplying by two pies. Let's 11.3 radiance per second. The moment of inertia of the turntable I sub t is 0.2 kilograms meters squared. The party that we're gonna drop on the term table has a mass m of 200 grams or 2000.2 kilograms. So we convert everything two kilograms here because we want to use S I units and the distance away from the center that the party has dropped his 20.15 liters. So for part A, we are asked to find the initial kinetic energy of the turntable. So since it's all rotational, the initial kinetic energy of the turntable which we call Katie I is equal to 1/2 Iomega initial squares. This is a make, I swear. So we just simply have to plug those values in when we find that the initial kinetic energy of the turntable is equal to 1.2 a jewels so we can box set in as their solution. For part a part B says, What is the final rotational speed of the system with the lump of the putty now on the turntable? So to do that, we're gonna consider the fact that the end of their momentum is gonna be conserved to the final angular momentum. Elsa, death is equal. The initial angular momentum will The final angular momentum is going to be the total moment of inertia of the putty plus turntable system. So this is the moment of inertia of the turntable I 70 plus m r squared to get the moment of inertia of the putty. Uh, where m is the mass and R is the distance away from the center that the putty lies and then omega f is the final rotational velocity. What we're trying to find this is equal to the initial moment in the momentum of the turntable, which is I so tee times, omega chi. So we know everything here except for omega F. So we simply saw for Omega death we find that this is equal to I sub t times Omega chi divided by Isom t plus him are square. So simply plug these values into this expression and we find that the, uh there. Um, rotational speed in the end is 9.23 radiance per second. So we can box that in is our solution for B Heart C has asked us to find the final linear speed of the lump of putty. Okay, So defined the linear speed of the lump of putty. You simply have to use the definition that v final. The linear speed is equal to go make a final times the distance the putty is away from the centre are so just play these values in and we find that this is equal to 1.38 meters per second. Okay, so now for the next three questions D, e and F, what we're going to be doing is we're going to find the change in kinetic energy of these different parts of the system. So first, we're gonna find the change in kinetic energy of the turntable. Okay, so this is, um, parte de we'll call this Delta Katie for the change in kinetic energy of the table. So this is just gonna be Katie final minus Katie initial. But we already found we go up. We already found Katie. Initial right so we just need Katie final. So this is gonna be Katie Final is gonna be 1/2 Iomega final squared where I is the moment of inertia of the table. We could also write out the expression for Katie initial, but we already have a numerical value for Katie initial cause we found that in part a party. So we can just write this again as Katie I because we already have that numerical value. Otherwise, you could write it as 1/2 isom t omega initial squared. So plugging these values in, we find that Delta Katie is equal to negative 0.43 jewels. So it actually loses some energy, which makes sense because it's been slowed down due to the putty being dropped on it. Which is why we have that negative sign. Okay, now, party were asked to find the change in kinetic energy of the putty. We'll call this Don't decay, P. Okay, but this is equal to K p f minus k p i. But the party had no initial energy, so that goes to zero. So this is just KPs So K p f is 1/2 times the moment of inertia of the Putty, which is just m r squared times that and your speed of the putty in the final state Omega F Square. That's the same Omega value is the turntable since their rotating at the same speed. So playing this in, we find that this is equal to 0.19 jewels. So it actually gains kinetic energy, Which makes sense because it started out with none. And now it's rotating. So it has some. And then lastly for F, we're asked to do it for the entire system. So, um, so for the entire system, we'll just call this don't decay. This is gonna be K final minus k initial. Okay, so que final minus minus K initial. Well, what is K final minus K initial. So this is K for Final is que t final plus KP final minus Katie initial because there is no KP initial. So this comes out to be equal to negative 0.24 jewels so we can make a little statement about this so we can say that the the total kinetic energy there is a total change in kinetic energy, which we're going abbreviate with Katie in this system is equal to the sum of the kinetic energy changes in the individual constituents that make up the system. It's a little bit bigger and weaken box all of that in as our solution to Part F, which is the final part of the question.


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