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When its KE is half of its maximum value how much PE does it have?
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juo ji bhar ke!
Three friends hotal main lunch kerny gy aur bill 75 rupees bana 3 dosto nay 25 25 25 rupess dy diy then hotal manager ny 70 rupess rhak ker 5 rupess waiter ko wapis dy diy so waiter nay 1 1 1 rupea 3?
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gweoi hwio ieiesjwio ueie ioeioe ihseie biwe ke iowe jdopejdie hieeuejje hyeoejieioe ioioewoi ooieuie ieoeieioe heie oeieieie odeueie uieioe ioope iodierieir hdjue bkdioeeioe iodioe iueioe oieid uehveie kwiioe eioeiow ueiepo oeioe iheoiod ieioe eioeoeoieue ieioe oieioe ieio ieieiie iw gwow udo i bwioe ioiowope jeieiow ueiow ieioe eiowow heope opeopeioe poeoi ueueie eior beiue oprope ideie iiweeieheioe heojeudioe iebeieuebeiooe jeioe ideneoe beoe iegeioe iowioe vepe ieioe iieioe ieeioejieke jekoeioe opeioe heioe0e heide ohdoe udioer uedie uieioe obeioe ioeoiioe oieope ioeujeie yeue eoe ioeikie ngoer ioeop4 prieioobeioee jkeioe beoeeioe ujeioeeioe je4io4vei4 iogeopyhe opeoe ioioe beio4 ioebgeior igeiowioe koe iooweioe ueir hediheoppeid idbiod jdio euei ioepoe ideie iror uisioe ugeoew ewioe iweoweioe jeioeioe ioeoe jeueeie ioeioe ieioeope ioeioe eiorioiio rioepore
What is the total energy of a projectile if its PE at the apex is 1050j and that value is twice half the maximum kinetic energy of the projectile at landing?
This is a trick question with the 'twice half' words. The energy is 1050 J which is PE at the apex, it is travelling vertically, there is no air resistance, and that energy has been totally converted to KE when the projectile hits the ground.
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Ke$ha's nationality is American. She was born in L.A, California in the United States.
How are potential energy and kinetic energy transferred on the first hill of a roller coaster?
At the top of the first hill, PE is at its maximum, whereas KE is zero. When the train starts to fall down the first hill, PE decreases and KE increases. At the bottom of the hill, KE is at its maximum, and PE is zero.
What happened to kinetic energy if the magnitude is being halved?
Kinetic energy = 1/2 M V2 .Double the mass . . . doubles the KE.Cut the speed in half . . . reduces the KE to 1/4 .Do both . . . reduces the KE to 1/2 its original value.
Why time recorded when half of sulphuric acid ran into reactant mixture?
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Which describes the relationship between potential and kinetic energy when a pole vaulter's pole has reached its maximum bend point?
When a pole vaulter's pole reaches its maximum bend point all the energy if the system is being stored as potential energy in the pole. Etotal = PE + KE. At this point KE is 0 so PE=Etotal. However, as the pole straightens out again, the stored PE is released as KE.
how is mechanical energy is related to roller coasters?
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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When a car's speed is reduced to half how is its kinetic energy affected?
KE = (0.5) x mass x V x V [at velocity V] KE = (0.5) x mass x (0.5V) x (0.5V) [at velocity 0.5V] Ratio = [0.5 x mass x V x V] / [0.5 x mass x 0.25 x V x V] = [1] / [0.25] = 4 This means that the KE will be 1/4 at half the speed compared to KE at the original speed
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How do the mass and speed of an object affect the energy?
Kinetic energy is the mass times one half the velocity squared. KE = ½mv².
