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E=k+U Ener gla MecanicaConservative ForcesNonconservative ForcesK=imV2 Energla Cinetlca U = Energia PotencialGravitational Elastic ElectricFriction Air resistance T...

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E=k+U Ener gla MecanicaConservative ForcesNonconservative ForcesK=imV2 Energla Cinetlca U = Energia PotencialGravitational Elastic ElectricFriction Air resistance Tension in cord Motor or rocket propulsion Push or pull by personUg = mgy Energia Potencial GravitacionalUs =Zkx? Energia Potencial Elistica En presencia de Fuerzas conservativas Ia Energia Mecanica del sistema se conserva (Se mantiene constante;all potential energyProblema #2: Si dejamos caer una piedra de m = 1.5 kg desde una altura

E=k+U Ener gla Mecanica Conservative Forces Nonconservative Forces K=imV2 Energla Cinetlca U = Energia Potencial Gravitational Elastic Electric Friction Air resistance Tension in cord Motor or rocket propulsion Push or pull by person Ug = mgy Energia Potencial Gravitacional Us =Zkx? Energia Potencial Elistica En presencia de Fuerzas conservativas Ia Energia Mecanica del sistema se conserva (Se mantiene constante; all potential energy Problema #2: Si dejamos caer una piedra de m = 1.5 kg desde una altura de h = 1.90 m. Determine la energia cinetica, la Energia Potencial y Ia Energia Mecanica. Complete la siguiente tabla. half U. hall K all kinctic encrgy K Ug Velocidades (joules) (joules)_ (mls) Punto (joules) B



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64. Integrated Concepts
(a) Assuming 95.0$\%$ efficiency for the conversion of electrical power by the motor, what current must the $12.0-\mathrm{V}$ batteries of a 750 -kg electric car be able to supply: (a) To accelerate from rest to 25.0 $\mathrm{m} / \mathrm{s}$ in 1.00 $\mathrm{min}$ (b) To climb a $2.00 \times 10^{2}-\mathrm{m}$-high hill in 2.00 $\mathrm{min}$ at a constant $25.0-\mathrm{m} / \mathrm{s}$ speed while exerting $5.00 \times 10^{2} \mathrm{N}$ of force to overcome air resistance and friction? (c) To travel at a constant $25.0-\mathrm{m} / \mathrm{s}$ speed, exerting a $5.00 \times 10^{2} \mathrm{N}$ force to overcome air resistance and friction? See Figure 20.47.

Hi. For this problem, we have a rocket with mass 2150 kg moving with with an impulse. But in this case, the impulse is given by a changing force in time with a T Square. Now, something that is provided in this with this equation is, at a time 1.25 seconds. The forces you go to 781.25 m for this problem were asked to find the constant A in the formula of the force. We're trying to find the impulse or 1.5 seconds after 200 seconds. And finally the change of velocity during this time interval the equations that we're gonna use today. It's impulse as the interval of the force when it changes to time and imposed. It's also equal to the change of momentum. With that in mind, let's begin. So we have the force at 1.25 is equal to 781 in 2030 mutants. In this case, let's plug in 1.25 in the original formula or in the formula with time. When we get that, uh, you can substitute this one on the left side. Who is this? Five is equal to eight times time, 1.25 square. So if I wanna find hey, and it's gonna be a 781.25 divided by 1.25 square. This one give us Kouadio 500. And in this case, since it has to make sense with force and this one is gonna be mutants over second square time giving seconds in square is gonna be second square times A We're going to get the units off Newtons of force. So this is the constant A with its units now, knowing that the force has this new form 500 I'm gonna not. I'm going to skip the units for now, but it's gonna have 102 square. I can use the impulse us the integral off a function, of course, in time, the prospective time to find the impulse. In this case, since we're a time 200 we need to find it at 200 plus the little time in this case, we're going from 200 to 201 5, 200.5. All right, so let's find it in Israel, in this case, 200. I'm good. 15 And it's gonna be like 100 t Square. It's where fighting. Yeah, We use customers here. We're gonna get the 500 is a constant. So put it outside. Divided by the three. Coming from the integral right to plus one is 3 to 3. And then we're gonna evaluate the cynical from +21 201.5 to 200 seconds. But so this is just calculus. The fundamental theorem of calculus tells me that the upper limit is subtracted by the lower limit. So that impulse is gonna be equal to 500 300 affect race it. I think that I'm gonna stop to my limits. 3.52 minus 100 Cure. All right, this is just a fundamental clear enough calculus. So they impose ones that we put in these families. We're gonna get on temples of 300 to to fine 962.5 kilograms kilograms, meters per second. Right. This is the people's Yes. Finally, we're asking to change the to find the change of velocity during this interval of time. So I remember the formula that I provide you impulse. It's also related to the change of momentum. Right now, the change of momentum is they go to the momentum one or final. Let's say one minus zero. In this case, I'm gonna take the form off. Um, mass, be one. Must be zero. But since I know that the mass is not changing, this one can take the form of mass times the one minus B zero No. One that the mass is the same. So this is the change of philosophy. So we're here. The change of momentum. I considered us mass times. The change of velocity and the change of momentum is the impulse. So we can say that impulse, this mass time change of velocity when the mass is not changing. So we take our previous value trees Well, let's let's software the change of velocity as the pimples divided by mass I'm in the masses provided 2001 15 That's passed this humble 312 It's beautiful, it's 2.5, but by the mask my men these kilograms meters were still can the programs. So you consider the Kilogram star? Yeah, Council. So we got a change of velocity of 14 Tuscan, 58.40 meters per second. And this is very changeable. Lost. Okay, Thank you for watching.

Problem. 8.11. We have a poorly drawn rocket in outer space with a massive 2150 kilograms. And at time T equals zero. It starts firing of an engine that produces a force that's proportional to time squared. We're told that the value of this force at one and 1/4 seconds after it starts firing is 781.25 moons. So we have several things were supposed to find, the first of which is what is the numerical value with units of A. So if we know that Oh, if we know T and F, then you know, we just solved this for a it's going to be 50. I didn't buy squared. So this is 700 81 a quarter Britons, divided by one and 1/4 seconds squared. And so this works out P 500. Newton's her second square. You can already tell what the units need to be. Uh, just this, you know, this has to have be have units of Newton's. This is gonna have units of seconds squared. So this needs to impudence of Newtons per second squared. So this this the unit's work out what you would hope And this is this is the number you get apart me is a little bit more interesting. It asks us to find what is the impulse that the this force produces? Does it exert on the rocket during the 1.5 2nd interval starting at two seconds after the T equals zero with the engine starts firing. So we know the J is equal to the after girl between two times of the force in this case to you, one is two seconds and he too will be three and 1/2 seconds. And so, putting in 80 spared here A doesn't end on times that comes out of the integral. That's medication Don't change. She's very t and just from Diggle, rules for polynomial is we know that the integral of T squared is 1 30 cubed. This has to be evaluated between are too times me and this ends up being 5008 110 Newton seconds, which is your recall. The impulse is also equal to the change in momentum. And so, you know, Newton seconds has the same units. If you expand out what Newton is a kilogram meters per second squared. It ends up having units of kilogram meters per second. So this is the same. It's the units for a moment now are at sea. What we want is to find what is the change in the rocket speed? Uh, because all of this is about the change in the momentum and everything. The actual initial speed of the rocket doesn't really matter. Before, there were no forces acting on it. So it was in, you know, an inertial reference frame. And, you know, you could look at it from whatever inertial reference for have you like and it wouldn't have changed anything, So I don't care what the speed was initially. Uh, admit that zero. It could have been whatever. So we know, Like I just said that the impulse is equal to the change in momentum. And now the problem says to assume that the mass, his constant of the rocket so it doesn't change from this 2150 kilograms, which is not how real world rockets work, but it doesn't make our lives a little bit. He's here here because now M doesn't change. And so this is going to become him Delta V and this is what we're looking for. So change in the speed is going to be equal to the change in momentum or or the impulse which you've already found divided by the mass which patrol. So you know, you have 5000 810 in seconds here, 2150 kilograms down here and this works out to be 2.7 meters per second. And you can see that even though it's, you know, a fair amount of force that this is exerting over quite a while. Um, because the mass of the rocket is so large, it doesn't speed up a lot in this one and 1/2 2nd interval, and that makes you know that makes intuitively.

The energy density ad Gasoline cheese 126,000. Barrister, my unit must I eat with 105 40 jewels. Now that's in gathered. So we have two divided by 3.786 times 10 to the minus three to get it in. Cubic meter on now, divided by the density. So 6 70 So finally you got it in your parking program. Now it would come out of the 5.24 times changed the seven use bar. You know, Graham, And then you density off battered E. It was Who has June Park? Awesome times 100 bosom times on our is 3600 seconds divided by its weight. So it is 2.7 times 10 to the five. You would bar kilogram. And the density of a capacities. Yes, Cities where CE 0.1 leads to a square divided by the sweat is 0.1 kilogram. So that was about to be 70 to do the ball, you know grand. So energy of the gasoline is highest

Hello, friends. This is the problem based on energy supply for in automobile. By using the arbitrary and a cap pistol by using the given data, we have to compare toe. Compare the energy per unit mass off energy source or gasoline you you with lead as it were. Tree and cap pistol. It is given that for gasoline commence a rate is 126000 We at you for density is given 6. 70 kg per meter. Cubed potential off lead. Acid battery is 12. World current capacity is 100 MPs over Mass is 16 k g. Potential difference at fully charged Napster is 12. Vote cap instances 0.1. For me on moss is 0.1 kg. Let us see you for gasoline. We can write energy is given 1 to 600 We get you per gallon. That is 1054 Jewell per Viet you, but gallon is called toe. 3.786 Tend to the poor ministry meter cube density, but Meter cube. It's called toe 6. 70 kg, so you will get 5.24 10 to the power seven Jewell per cages. So it's convention rate is 5.24 and toe 10 to the power seven Jewell per k G. Now for battery, it is well, Jewell per cola 100 too long per second in two seconds. 3600 seconds upon 60 kg. So it is to be 2.7. Uh huh. In tow 10 to the power fight Jewell for K G on for cap Easter It is half energy off the capital studies half see Delta v the cap instances 0.1 potential differences 12 volt upon masses 0.1 kg So 72 door for K G. So if we compared a specific energy off gasoline upon a specific energy off battery and cap pistol, then it is Toby 5.24 10 to the power seven. Upon 217 and 10 to the power five plus 72 it will be I don't 7 to 7 Jiro Jiro times much. That is the answer for it. Thanks for watching


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