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"fo ke je te parle" is text language for "il faut que je te parle" meaning "I need to talk to you".
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Imagine you have a roller coaster which starts moving from point A down to point B, which is at ground level (where height, h, is equal to zero). It then moves up to point C, which is at about half the height of point A, then down to point D, which is slightly above ground level. Then it moves up again to point E, which is at a greater height than point A, and in doing so passes point F, which is at the same height as point A (drawing this out will help or look at the related link below for a diagram).
TE=total energy
PE=potential energy
KE=kinetic energy
Assuming friction and air resistance are negligible and that the roller coaster starts from rest, then the TE of the roller coaster is equal to its PE at point A.
TE=PE at A
As the roller coaster moves from A to B, its PE changes into KE. Since h=0 at B, then all the PE of the roller coaster at A is turned into KE at B.
The change in PE=the change in KE from A to B.
Here it is useful to note that at A, KE is a minimum (0) and PE is a maximum; at B, KE is a maximum and PE is a minimum (0). Thus, the KE at B is also equal to the TE.
TE=KE at B
Also note that TE remains constant, being the sum of the PE and KE possessed by the roller coaster.
PE at A=KE at B
At A, TE=PE+0
At B, TE=KE+0
Hence, TE is constant.
As the roller coaster moves from B to C, its KE changes into PE as its height above the ground increases. However, when it reaches C, it does not possess only PE, but a combination of PE and KE.
TE at C=PE at C + KE at C
The reason why PE is not a maximum at C is because C is lower in height than A. We know that PE at A is the TE of the roller coaster for the entire course. Since PE is dependent on height, in order for the roller coaster to reach maximum PE, it must be at a height equal to the starting height. C is at roughly half the height of A, hence the roller coaster will possess only about half the PE it had compared to when it was at A. The rest of the energy is KE since TE=KE+PE.
D is not at the same level as B, but is slightly higher. Hence, the roller coaster will not move as fast at D than it did at B. This is because it has less KE at D, due to the fact that it still possesses some PE (since h is not equal to 0 at D).
Since TE=KE+PE and PE is not equal to 0, then KE will not be maximum and thus the roller coaster will move less quickly at D than it did at B.
Using the same principle, the roller coaster will not be able to reach E. This is because it reaches maximum PE when it is at F, since F is at the same height as A. We know that at A, PE=TE. Hence, at F, PE=TE. Energy can neither be created nor destroyed, hence the energy of the roller coaster cannot exceed the TE it had at the start. Therefore, it will not reach E, but it will be at rest momentarily at F before moving down again and back to A (remember friction and air resistance are negligible), and continue moving back and forth between A and F.
However, the roller coaster will be able to reach E if it is given KE in addition to the PE at A. In other words, if the roller coaster is already moving at a sufficient speed as it passes A, then it will be able to reach E. This is because the TE at A will now be equal to the sum of KE and PE at A, and KE is not equal to zero as it was in the previous example. The additional KE that would need to be supplied in order for the roller coaster to reach E would be equal to the difference in the PE at E and the PE at A (or F).
PE at E - PE at A = KE at A
which is the same thing as
TE - PE at A = KE at A; or TE=PE at A + KE at A
That's pretty much all of it.
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Bukina ba ana ongora ke ana kakaea raoi te koaua ba antai ae beka iaon te bike
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Quedar is the form 'To stay', therefore you stayed would be quedaste
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toi muon tin ke hoach nha hoc duong. che do danh cho cong tac vien kiem nhiem y te hoc duong nhu the nao?
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The motto of Hamilton East School is '"Tino nui ki ahau ke te ako"'.
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"You are (literally) doing yourself/becoming/growing."
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bakwaas na kar penchod lan te charaa ke nachau painchoda
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The type of engine that can be built into a 97 Cadillac is the V8, 4.6 Liter, Transmission.
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Chupa is a term that can refer to various things, but it could be related to the Chupa Chups brand of lollipops, or it could refer to a type of mythical creature like the chupacabra in Latin American folklore. Can you provide more context or specify which type of "chupa" you are asking about?
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Mingqing. Ke has written:
'Zhong cao yao you xiao cheng fen li hua yu yao li te xing' -- subject(s): Materia medica, Pharmaceutical chemistry, Pharmacognosy, Pharmacology
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ture ke o cha pinda c us kanger ne 101 alcohol ke bottles p hai ye to hamare leader he nahi anuh te liter maro
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We start with the fundamental tenet that all closed system objects have some total energy TE that can be neither created nor destroyed. Which is to say TE = constant as long as we are not adding energy to it like through work or taking energy out of it like through friction.
So all objects can have a TE = PE + KE + QE = constant; where PE is potential energy, KE is kinetic energy, and QE is all other energies, like friction heat or work done on the object. So let's see what happens to the RC cart at its various waypoints along the tracks.
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While it's picking up passengers, TE = PE + KE = 0, the cart has no energy. It's just sitting there not moving, waiting for you to climb aboard. And it's at the lowest point on the tracks; so there is no potential energy.
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Now, as you are safely aboard, the cart is pulled up by the force F of some motor pulling it up to the (gasp) highest peak H on the track. So work energy QE is added to TE and we now have TE = QE = PE because all that work was transformed into potential energy TE = PE = mgH where m is the mass of the cart and you at the height H.
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And over you go. At the peak H you and the cart have TE = PE = mgH potential energy, but for a moment no kinetic energy KE. But discounting friction losses, now that H is getting smaller as you plunge toward a low point h < H in the track the potential energy drops from PE = mgH to pe = mgh < mgH = PE. And there you are, since TE = constant = PE + ke if the PE is getting smaller, pe < PE, the kinetic energy ke must be getting bigger KE > ke to keep TE constant. That means the cart's speed is increasing as you fall to the bottom of the track's height. To use your term, the PE has been transferred to KE by the drop from that height H.
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But you survive that breath taking plunge and because the cart has lots of KE at the bottom, it continues on past the h point and starts to climb back up to yet another high point. And because TE = KE + pe = constant as the cart and you go back up you are converting that kinetic energy into a new potential energy level; so at the second high point you have TE = PE and virtually no kinetic energy at the new peak. All this repeats for every plunge along the way.
In sum, the conversions (transfers) go something like:
TE = 0 + QE = PE + ke = pe + KE = PE + ke = ... = constant. Where we see the potential energy at each high point is traded off for kinetic energy at each low point. But, overall, the total energy stays constant when friction is discounted.
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Quel type de garçons te fait craquer? in French is "What kind of boys attract you?" or "What type of boys do you fall for?" in English.
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Assuming the simplest conditions (ie (1) the pendulum pivot is frictionless, (2) no air resistance for the bob's travel, etc...):
This is a problem best solved by considering the situation from an ENERGY standpoint (and not from, say, a velocity and force standpoint).
Under energy conservation (the assumption of "simplest conditions" above gives us this), the total energy (TE) remains constant. The two components contributing to the total energy in this situation are: (1) the kinetic energy (KE) and (2) the potential energy (PE).
So, TE = KE + PE
We know (hopefully) that the (simple) formula for the PE of a mass in earth's gravity is:
PE = mgh where m is the mass, g is the force of earth's gravity and h is the height above SOME (any) reference point.
We also know that the pendulum's KE at the top of its swing (in this problem, the top of the swing is when it's horizontal) is zero.
For simplicity, we'll set the reference point for all height measurements at the bottom of the pendulum's swing (ie one meter below the pendulum's point of attachment to the stand/ceiling/support whatever).
Putting all the info. above together:
At the START, TE = PE + KE = PE = mgh = 0.1 kilograms * 9.8 m/s^2 * 1 meter
= 0.98 Joules
Now at the BOTTOM of its swing, TE = PE + KE,
BUT PE is zero at the bottom of the swing (because h is zero)
AND we just computed TE above as 0.98 Joules
SO, at the swing bottom, KE = TE = 0.98 Joules
At the 30 degree from vertical mark,
Trigonometry tells us that the height, h, at 30 degrees is 1 meter - cos(30)*1 meter
= (1 - 0.8660) meters = 0.1340 meters
SO here the PE is: PE = mgh = 0.1 kilograms * 9.8 m/s^2 * 0.1340 meters
= 0.1313 Joules
AND going back to find the KE, TE = KE + PE gives us 0.98 = KE + 0.1313,
implying that KE = 0.8487 Joules
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Let's get started! An object in motion possesses kinetic energy and to bring the object to a stop this kinetic energy must be removed. ... A mechanical brake applies a friction force to convert the kinetic energy of the vehicle into thermal energy which then dissipates into the atmosphere.
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No. "Te quiero" is more passionate than the type of relationship a mother should have with her child. Better expressions would be "Te amo", "Te adoro", and "Me encantas" - but this is used more with things that they do.
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I don't understand either what country you are from
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Con Su Pinche Madre Chingados Segun Esto Te Tiene Le Dar las answers y ni te da nada la chingadera porke awevo kiere ke uno la ponga si no dise nada
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