OlegGorokhov

Discovery #2 The detection of Energy Imbalance. The New Energy. The New Ability of Electromagnets. About the Multipliers of Electrical Energy Imagine that there is a bank that gives you a free loan every day — the loan that you do not need to return. And this bank has only one condition : you need to have some own amount. Any amount. And this bank will multiply any of your amount 300 times. Imagined it? Now imagine that this bank would give you that loan 2 times a day. And now imagine that yet more often — every second. Imagined it? Would you like to have such a bank? And such a bank exists. This bank is an electromagnet. You can give an electromagnet some Energy (for example, from the battery) and the electromagnet will give you 300 times more Energy. This will be happening every second while you give the Electromagnet some Energy. Electromagnets simply work like Solar Panels and also like solar panels take Energy from the surrounding space. However, electromagnets, in contrast to solar panels, take some new Energy (i.e., previously unknown Energy) and can take this new Energy even from the dark. And now let’s consider a concrete example in detail and with numbers. First some words about how an electromagnet works. Any electromagnet indeed is just an ordinary coil of wire. The wire must be lacquered. Such wires are inexpensive and freely sold in every city and town. Here’s how any electromagnet works. You just need to connect the ends of the wire to any battery. That’s it. Any electromagnet can push closer to itself (attract) a iron object and hold it in the air motionless. We can always increase the lifting Force of an electromagnet by putting an iron rod inside the coil. So an electromagnet can create the lifting Force pushing up, which works against the gravitational Force pushing down (i.e., an electromagnet can do the same thing as you did, holding the 20kg object). Let’s take the most ordinary electromagnet. That electromagnet is always available on eBay, Amazon, Alliexpress and so on. It’s about $19. Now we already know that this electromagnet is just a coil of wire with a metal rod inside. So our electromagnet (bought on eBay) has these properties: 24V (24Volts) 0.34A (0.34Amperes) 50kg (50kilogram) This means that this electromagnet can hold in the air the 50kg iron object by consuming Energy at the rate of 1.66 Energies per 1 second. Here it is necessary to explain how the Energy consumed by the electromagnet per 1 second (E) is calculated. To calculate it, Volts have to be multiplied by Amperes or in other words, Voltage (U) has to be is multiplied by Amperage (I). E = U*I = 24(V)*0.34(A) = 8.16 Joules per 1 second We received that the electromagnet, holding 50kg object in the air, consumes Energy at the rate of 8.16 Joules per 1 second. But since the Joule is the erroneous unit of measurement of Work and Energy, then to get the actual data in new units (i.e., in Energies), we need to correct the Voltmeter’s readings, namely, multiply the Voltmeter’s reading by the correction coefficient (k). E= k*U*I Since there inside each Volt 1 Joule “sits”, then to calculate the correction coefficient, we must relate the new formula and the old formula by this correction coefficient. As the old formula we will use the formula for calculating the Static Mechanical Work in Joules. And now we calculate this correction coefficient (k). Substitute the relevant values: a = 9.8(m/s²); t = 1(s). k = 2/(a*t) = 2/(9.8(m/s²)*1(s)) = 1/4.9 = 0.204 And then we get, E= k*Voltage*Amperage = 0.204*24(V)*0.34(A) = 1.67 Energies per 1 second And so we received that the Electromagnet, holding 50kg object in the air, consumes Energy at the rate of 1.67 Energies per 1 second. Now let’s calculate the Mechanical Work that this Electromagnet is doing within each 1 second, when holding 50kg object motionless in the air or, in other words, the Static Mechanical Work that the pushing Force of the electromagnet is doing when pushing up the 50kg object for 1 second. The mechanical Work of the pushing (attracting) Force of this electromagnet here also equals the mechanical Work of the gravitational Force pushing down the 50kg object. So, calculating the Work of the gravitational Force per 1 second, we simultaneously calculate the Work of the pushing (attracting) Force of this electromagnet per the same 1 second. W (E)= 50(kg)*9.8(m/s² )*1(s) = 490 Energies in 1 second Thus, holding 50kg object in the air, this electromagnet creates the pushing Force, which per 1 second does the Mechanical Work equal to 490 Energies. Attention! 490 Energies per second is 300 times greater than 1.67 Energies per second. This means that every second this simple device does its Work which is 300 times more than the Energy this device consumes every second. Let’s calculate the Efficiency of the electromagnet. Efficiency = Work/Energy*100 Efficiency = 490(Energies)/1.67(Energy)*100 = 29 424% Thus, the Efficiency of this electrical device (this electromagnet) is more than 100% !!! is 29 424% !!! ______________ It’s important to understand that we get the same Efficiency (29 424%) even if we calculate the Energy and the Static Mechanical Work in Joules. Using this formula we can calculate the Static Mechanical Work in Joules. W (E) = 50(kg)*(9.8 (m/s²))²*(1(s))² = 2 401 Joules per second Efficiency = Work/Energy*100 Efficiency = 2 401(J)/8.16 (J)*100= 29 424% This is because of the fact that when the Efficiency is calculated, then the Work is divided by the Energy and it means that even the erroneous unit of measure (Joule) gives the correct result, since one error (the same coefficient) “sits” both in the numerator and in the denominator and this error (this coefficient) is reduced. ______________ So, the detection of this enormous Energy Imbalance means at least 3 things. 1. We have discovered some new Energy that magnets (electromagnets and permanent magnets) take from the surrounding space and turn it into the pushing (attracting) Force. (This comes from the following conclusion: If something can turn into one that can push other thing then this somethingis Energy). In other words, we have found a really inexhaustible source of clean energy. 2. We have discovered the new ability of magnets (electromagnets and permanent magnets) — the ability to take Energy fom the surrounding space. 3. We can multiply Electrical Energy. All this means that we can build the Multiplier of Electrical Energy (MEE) that multiplies Electrical Energy and works with the extremely high Efficiency. Let’s take a closer look at the Work of our electromagnet while holding 50kg object in the air. Every second the electromagnet does Work equal to 490 Energies, but receives every second from the battery just 1.67 Energies. The difference (490–1.67 = 488.33 Energies) the electromagnet takes from the surrounding space. Thus, every second the electromagnet takes 1.67 Energies from the battery and 488.33 Energies from the surrounding space and then converts both these Energies (in the sum of 1.67 + 488.33 = 490 Energies) into the pushing (attracting) Force. And this pushing Force every second does Work equal to 490 Energies . About the Multipliers of Electrical Energy Using Electromagnets and certain engineering solutions, we can create Multipliers of Electrical Energy, which multiply Electric Energy. And multiplymany times — hundreds of times! I think I know how to build the Multiplier of Electrical Energy — how exactly to take out this new Energy from electromagnets and turn this new Energy into Electrical Energy. I don’t wanna patent this idea of the Multiplier of Electrical Energy, instead I just wanna get the attention with my discoveries and then share my knowledge with people. After all, we can take Energy from the Sun and multiply it by the Multipliers of Electrical Energy. This will change the entire generation of Energy on the planet. Discovery #3: (temporally hidden) The New Law of Conservation of Energy. The New Concept of Charge. The New Concept of Mass. I’m looking for an informational sponsor for the presentation of this discovery. TheOlegGorokhov@gmail.com #Gorokhov ____________________ I hope these discoveries will also kick-start the process of total change of the whole world’s education system. I have ideas on it. So every child regardless of location will enjoy schooling and get the best education on the planet for free.

The biggest discoveries in physics Hello. I am Oleg Gorokhov. TheOlegGorokhov@gmail.com I made a series of the biggest and simultaneously simple discoveries in physics. These discoveries are so big that they completely change our understanding of the universe and give us access to new colossal clean Energy. This text is presented in such a way as to be understandable to everyone. You are able to understand this yourself. Trust yourself — your mind. Believe in yourself. Discovery #1 The New Concept of Work and the New Concept of Energy. The New Universal Formula for Work and Energy. The New Unit of measure for Work and Energy. The blind pushing and the Static mechanical Work. The New Formula for Kinetic Energy. The New Formula for Potential Energy. The New Formula for Power. The erroneousness of the Joule. In this section I proved the erroneousness of using the Joule as the unit of measure for Work and Energy and the erroneousness of the “current” formula for Work and Energy. The aftermaths of the Discovery#1 Discovery #2 The detection of Energy Imbalance. The New Energy. The New Ability of Electromagnets. About the Multipliers of Electrical Energy (MEE) Discovery #3 The New Law of Conservation of Energy. The New Concept of Charge. The New Concept of Mass. ________________ Also first you can read these links on Medium: https://goo.gl/aETJdQ (3 min read) https://goo.gl/YffRQL (5 min read) Preface What is Science? Science is not a set of dogmas. This is our attempt to understand the universe. Therefore, any concept or formula in physics is not a stiff dogma, but only our attempt to understand reality and describe it in words and in formulas. In physics, the concepts of Work and Energy are basic and, accordingly, are the bases for constructing other physical concepts. This is hard to believe, but throughout the entire history of science, people have used erroneous formulas to calculate Work and Energy and, as a consequence, the erroneous unit of measurement of Work and Energy — the Joule. As a result of these basic errors in science, most processes in the universe fell out of our vision, so humanity had an absolutely wrong picture of the physical world and was losing gigantic possibilities for all this time! Now my discoveries open these possibilities to us. Let’s start. _____________ Discovery #1 The New Concept of Work and the New Concept of Energy. The New Universal Formula for Work and Energy. The New Unit of measure for Work and Energy. The blind pushing and the Static mechanical Work. The New Formula for Kinetic Energy. The New Formula for Potential Energy. The New Formula for Power. The erroneousness of the Joule. In this section I proved the erroneousness of using the Joule as the unit of measure for Work and Energy and the erroneousness of the “current” formula for Work and Energy. The aftermaths of the Discovery#1 In my opinion, basic errors in science occurred because people have not fully understood all the three Newton’s discoveries: Newton’s 1st, 2nd and 3rd Laws. Before I start the explanation what the New Concept of Work and the New Concept of Energy are, one should fully understand the following key concepts:  — The Force;  — The target acceleration  — The blind pushing. The Force. The concept of Force was presented by Isaac Newton in the distant 1686 year and is known to us as Newton’s 2nd Law. Force (F) is a pushing an object that tries to increase the velocity of this object. In addition, we can say that Force (F) is the size of the pushing per 1 second. Sometimes the Force (F) can increase the velocity of an object but sometimes not, if this object is also being pushed by another (equal or greater) counter Force (F1). Here is this formula of Force. Where m — the mass of a pushed object, kg a — the increase in the velocity of the pushed object per 1 second (acceleration), m/s/ s or m/s² This formula is easier to understand through a simple example: a 20kg object is freely falling down from some height. The 20kg object is falling down only because this object is pushed down by the gravitational Force. The gravitational Force (F) can be imagined as a pushing by “the invisible hand”. The green arrow shows the direction of this pushing. Before starting the falling, the velocity of this 20kg object was 0 (m/s), then this 20kg object begins to fall down and the velocity of this object is constantly increasing uniformly. At the end of the first second of the fall the velocity becomes 9.8 (m/s), at the end of the second second — 19.6 (m/s), at the end of the third second — 29.4 (m/s), and so on. For every second, the velocity of this 20kg object is evenly increased by 9.8 (m/s). The Force (F), the gravitational force, pushing down this 20kg object is calculated as the product of the mass of this object (m) — 20 (kg) and the increasing of the velocity of this object per 1 second — 9.8 (m/s). Or, in other words, as the product of the mass of this object (m) — 20 (kg) and the acceleration of this object (a) — 9.8 (m/s²). F = 20(kg)*9.8(m/s²) = 196 Newtons So, the gravitational force pushing down this 20kg object is equal to 196Newtons. In other words, it can be said that the size of the pushing per any one secondof falling is 196 Newtons. The target acceleration. The target for the gravitational Force is to give any object the down acceleration of 9.8 (m/s²). Therefore, we can call this down acceleration of 9.8 (m/s²) the target acceleration for the gravitational Force, the target acceleration for this Pushing. However, the gravitational Force does not always manage to impart this target acceleration of 9.8 (m/s²) to an object, although the gravitational Force always tries to do this by pushing each object down. For example, when the 20kg object is lying motionless on the table, the gravitational Force (F), pushing this object down, tries to impart a target acceleration of 9.8 (m/s²) to this 20kg object. But this can not be done, because according to Newton’s 3rd Law, the table (the table Force, F1) is also really pushing this 20kg object, but pushing it up, trying to give this 20kg object the same (in magnitude) target up acceleration of 9.8 (m/s²). Just think about it again — according to Newton’s 3th Law, the table (the part of the table surface) is also really pushes this 20kg object (exactly like the gravitational Force, F) and tries to accelerate the object! As a result, these two equal counter Forces (these two real counter Pushings) balance each other and only because of it this 20kg object remains stationary, which means that the visible accelerations of this object in both directions (up and down) equal zero. In such situations, when the visible accelerations of an object were equal to zero, people, when calculating the Forces pushing this object, always correctly used the target accelerations for these Forces. So using the target down acceleration for the Gravitaional Force (a=9.8 m/s²), we can already calculate the gravitational Force (F) pushing down this 20kg object. F = 20(kg)*9.8(m/s²) = 196 Newtons Thus, only using the target down acceleration for the Gravitaional Force, we get that the gravitational Force pushing down this 20kg object is equal to 196 Newtons. As already said, according to Newton’s 3rd Law, the table (the table Force, F1) is also really pushing this 20kg object, but pushing it up. This table Force (F1) is equal to the gravitational Force (F) i.e. equal to 196 Newtons. F = F1 = 196 Newtons This means that in calculating the table Force (F1) pushing this 20kg object up, we also use the target up acceleration for the table Force (F1), which is 9.8 (m/s²). F1 = 20(kg)*9.8(m/s²) = 196 Newtons Thus, only using the target up acceleration for the table Force (F1), we get that the table Force (F1) pushing up this 20kg object is equal to 196 Newtons. Next, we will also use the target accelerations in calculating real Forces! The blind pushing. The gravitational Force can be also understood by us as a global device pushing objects. Let’s consider 2 examples at once:  — the global device (the gravitational Force, F) pushes the 20kg object that is free-falling down;  — the global device (the gravitational Force, F) pushes the 20kg object that is lying on the table. In each case the global device (the gravitational Force, F) pushes the object with the same effort equal to 196 Newtons. I consider that during the pushing, the global device (the gravitational Force, F) doesn’t even “understand“ whether the 20kg object is moving from this pushing (the object is free falling) or not (the object is lying on the table) — in each case the global device just pushes this 20kg object with the same effort equal to 196 Newtons. So if there is a possibility for the object to move (i.e. the object is free falling), then the gravitational Force pushes the object and the object moves. If there is no possibility for the object to move (for example, the object is lying on the table), then the gravitational Force also pushes the object and the object is just pressed to the table surface. In both those cases the gravitational does the same thing — it pushes a object with the same effort equal to 196 Newtons. We can call this process (when the gravitational Force doesn’t even “understand“ whether an object is moving from its pushing or not) the blind pushing. This concept, the blind pushing, is very important and works like the Butterfly effect. It changes everything in physics. We will see it below soon. The New Concept of Work and the New Concept of Energy. I consider that in the most general sence: Work is any activity that requires effort; Energy is an ability to do Work. In other words, Work is any activity that requires Energy (or Work is any activity Energy is spent on). Then in physical terms: Work is a pushing an object for some time; Energy is an ability to push (an object) for some time. Some Works, at first glance, may not look like pushings. But this is only at first glance. Any Work is always just a pushing for some time or a combination of several pushings. I will give one example. A heating-up is also the result of pushings, because a heating of a body is due to the increase in the velocities of particles inside the body. And the increase in the velocities of particles occurs only due to additional pushings of these particles. Therefore, a heating-up is the result of a lot of pushings. And this means that a heating-up is Work and that Energy is being spent in this process. There are the Target Works and the Non-target (but inevitable) Works. The Energies that are spent on the Non-target (but inevitable) Works, we always consider our losses. For example, the Target Work of the electric motor is to push the motor shaft. At the same time, the heating of the electric motor is the Non-Target (but inevitable) Work. The Energy that is spent on this Non-Targeted (but inevitable) Work is regard as our losses. It is important to understand that when any Work (any Pushing)is being done, Energy is always being spent on this Work (on this Pushing). Always. When some closed system (for example, some device) pushes an object, simultaneously the ability of this system (this device) to push this object decreases. If Energy is spent on something, then this something is Work. If something can push other thing then this something is Energy. If something can turn into one that can push other thing then this something is Energy. The New Universal Formula for Work and Energy. Work (W) is the pushing an object with a Force (F) for a time (t); Energy (E) is the ability to push an object with a Force (F) for a time (t). Although this new formula of Work and Energy looks like the formula for Momentum (p), in fact, the concepts of these formulas are totally different. All this will be explained in detail several sections lower. Inserting the product (m*a) instead of the Force (F) in W(E)=F*t, we get the expanded Formula for Work and Energy. Now, this is the New Universal Formula for Work and Energy. The New Unit of measure for Work and Energy. I propose to name the new unit of measurement for Work and Energy the Energy. The dimensional formula of the Energy is: Energy = Newton*second = kilogram*meter/second And now let’s calculate the mechanical Work that the gravitational Force (F) does for 3 seconds pushing the 20kg object free-falling down. W (E) = 20(kg)*9.8(m/s²)*3(s) = 588 Energies Thus, pushing 20kg object for 3 seconds the gravitational Force does the Mechanical Work equal to 588 Energies. The blind pushing and the Static mechanical Work. Let’s once again consider the 2 familiar examples:  — the global device (the gravitational Force, F) pushes the 20kg object that is free falling down;  — the global device (the gravitational Force, F) pushes the 20kg object that is lying on the table. The Work of the global device (the gravitational Force) in each case consists only in a pushing the object with the same effort equal to 196 Newtons. Doing this Work the global device (the gravitational Force), as we know, doesn’t even “understand“ whether the 20kg object is moving from this pushing (the object is free falling) or not (the object is lying on the table). I consider that since the global device (the gravitational Force) in both those cases does the same thing — it pushes the object with the same effort equal to 196 Newtons — then in each case the global device (the gravitational Force) does the same Work and spends the same Energy on this Work for the same time. And now let’s calculate the Work of the gravitational Force (F) for 5 seconds for both those cases. First, let’s calculate the mechanical Work that the gravitational Force (F) does for 5 seconds pushing the 20kg object free-falling down. W (E) = 20(kg)*9.8(m/s²)*5(s) = 980 Energies Thus, pushing 20kg object for 5 seconds the gravitational Force does the Mechanical Work equal to 980 Energies. And now, let’s calculate the mechanical Work that the gravitational Force (F) is doing for the same 5 seconds pushing the 20kg object lying on the table. W (E) = 20(kg)*9.8(m/s²)*5(s) = 980 Energies Thus, pushing 20kg object for the same 5 seconds the gravitational Force does the Mechanical Work equal to the same 980 Energies. Now in physics there isn’t any term for this Work. We were told that in order to do the mechanical Work there must always be a displacement of an object. You yourself can see this by opening this wiki link. https://en.wikipedia.org/wiki/Work_(physics) So I introduce an absolutely new concept to science. I introduce the Static Mechanical Work to physics, to life. The adjective Static here means moveless. It’s a very important concept that changes everything in physics. It also completely changes our understanding of the universe. So now we can call this Mechanical Work, that the gravitational Force (F) does, the Static mechanical Work or, in other words, the Moveless Mechanical Work. W (E) = 20(kg)*9.8(m/s²)*5(s) = 980 Energies Thus, pushing 20kg object for 5 seconds the gravitational Force does the Mechanical Work (the Static mechanical Work) equal to 980 Energies. It is important to understand that always when a device pushes an object, it means that this device does the Work and the Energy of this device is spent on this Work, regardless of whether the pushed object is being moved from this pushing or not, for example, when some device pushes a fixed wall, then the Energy of this device is being spent on this pushing, despite the fact that the wall can remain moveless. __________________ Also it is important to understand that there is even an opportunity to calculate the Static Mechanical Work in Joules! We could use this formula to calculate the Static Mechanical Work in Joules: where, W (E) — Work (Energy) (kg*m ²/s² or Jouilly); m — the mass of the pushed object (kg); a — the acceleration of the pushed object, sometimes the acceleration can be invisible, i.e. Target (m/s²); t — the pushing time of the object by Force (s). This formula is obtained by simple transformations of the following known formulas: But we can not use this formula to calculate the Static Mechanical Work, since the Joule is an erroneous unit of measurement for Work and Energy! All this is explained in detail in the “The erroneousness of the Joule” section below. _____________________ As we know, when the gravitational Force (F) pushes the 20kg object lying on the table, the table (the table Force, F1), according to Newton’s 3th Law, also really pushes this 20kg object (exactly like the gravitational Force, F) and tries to accelerate the object! And now, let’s calculate the Mechanical Work (the Static Mechanical Work) that the table (the table Force, F1) is doing for the same 5 seconds pushing the 20kg object up. W (E) = 20(kg)*9.8(m/s²)*5(s) = 980 Energies So, pushing up 20kg object for 5 seconds the table (the table Force) is doing the Mechanical Work (the Static mechanical Work) equal to the same 980Energies. So, the Work of the table (the Work of the table Force) pushing up the 20kg object is equal to the Work of the gravitational Force pushing down the 20kg object for the same time (5s). The Energy that the table is spending on its Work for 5 seconds (pushing up the 20kg object) is non-zero. This new understanding certainly breaks the existing Law of Conservation of Energy. The New Law of Conservation of Energy is presented in the Discovery #3. __________ If this new knowledge (that Energy of the table is spent without any movement of 20kg of the object) confuses you, just think about how many small particles are , for example, inside the iron table.These particles are pushing each other to stay together in form of the table. These pushings are Work, and, of course, Energy is wasted on this Work. Think about how much Energy is spent on these pushings every second, in order the table simply remains in the form of a firm table. And, of course, extra Energy is spent on these pushings when there is the 20kg object on the table’s surface to prevent the particles from dispersing apart from each other and to prevent this 20kg object from going downwards (this is what the 20kg object is trying to do every moment). ________ Let’s consider other example: you starts holding the 20kg object in the air moveless. Now you have started doing at least 2 new Works: the Targeted mechanical Work (the Static mechanical Work) on the pushing up the 20kg object with your Force (F1) and the Non-Targeted (but unavoidable) Work on the additional heating of your body. As already mentioned before, Energies that are spent on the Non-Targeted (but unavoidable) Works, we always regard as our losses. Now let’s calculate the mechanical Work (The Static mechanical Work) that you (your Force, F1) are doing for, say, 1 minute holding the 20kg object in the air motionless. As we already know, the Work of your Force (F1) pushing up the 20kg object is equal to the Work of the the gravitational Force (F) pushing down the 20kg object. And since we can easily calculate the Work of the gravitational Force (F) pushing down the 20kg object for 1 minute, we can also easily calculate the Work of your Force (F1) pushing up the 20kg object for the same 1 minute (60 seconds), because of the fact that these 2 Works are equal. W (E) = 20(kg)*9.8(m/s²)*60(s) = 11 760 Energies Thus, holding 20kg object in the air motionless, you, within 1 minute, does the Mechanical Work (The Static mechanical Work) equal to 11 760 Energies. Let’s consider a situation when a jet vehicle (for example, Falcon 9) is hanging motionless in the air for 15 seconds. The mass of Falcon 9 is equal to 549000kg. When the this device (Falcon 9) starts hanging motionless in the air, it means that this device creates the lifting Force (Thrust, F1) pushing up the 549000kg device which is equal to the gravitational Force (F) pushing down this 549000kg device and only because of it this 549000kg device remains moveless in the air. We will not consider other the Non-Targeted Works which this device (Falcon 9) does as well. Let’s calculate the mechanical Work (the Static Mechanical Work) that is done by the lifting Force (Thrust, F1) pushing up the 549000kg device for 15 seconds. W(E) = 549 000(kg)*9.8(m/s²)*15(s) = 80 703 000 Energies Thus, holding 549000kg object in the air motionless, the lifting Force (Thrust), within 15 seconds, are doing the Mechanical Work (The Static mechanical Work) equal to 80 703 000 Energies. So we go on. The New Formula for Kinetic Energy Kinetic Energy of a moving object is equal to the maximum Work that this moving object can do when it collides with another object. This Work, in the event of the collision, is a pushing other object and this pushing may result the following occasions:  — the new velocities of both the objects;  — the deformation (destruction) of both the objects;  — the heat-up of both the objects, etc. Since the velocity of any object (v) at equally accelerated motion is calculated as the product of the acceleration of this object (a) and the time of the movement of this object (t): v = a*t, then inserting the velocity (v) instead of the product (a*t) in the Universal Formula W(E) = m*a*t, we get the New Formula for Kinetic Energy of a moving object: W(E) = m*a*t = m*v. For example, let’s calculate the Kinetic Energy of 20kg object moving (falling down) with an acceleration of 9.8 (m/s²), say, at the end of the 7th second after the beginning of the movement (falling). First, we calculate the velocity of this 20kg object (v) at the end of the 7th second after the beginning of the movement (falling). v = a*t = 9.8(m/s²)*7(s) = 68.6 meters per second And then we calculate the Kinetic Energy (E) of this 20kg object at the end of the 7th second after the beginning of the movement (falling). W(E)=20(kg)*68.6(m/s)=1 372 Energies Thus, this 20kg object, falling down, at the end of the 7th second after the beginning of the movement (falling), has such Kinetic Energy that, in the event of a collision with another object, it can do the maximum Work of 1,372Energies. It should be noted that the new formulas of Work and Energy are very similar to the Momentum formulas (p). https://en.wikipedia.org/wiki/Momentum But although these two groups of formulas look the same, in reality, the concepts of these formulas are totally different. The Momentum of an object (p) has always been understood solely through the visible movement of the object, and therefore only through the visible velocity and the visible acceleration. Therefore, scientists consider that if there is no visible movement of an object (i.e., the visible velocity and visible acceleration of this object are zero), then the Momentum of this object (p)is also zero. While the new concept of Work and Energy works, even when there is no visible movement of the object, i.e. when the visible velocity and visible acceleration of the object are equal to zero. The New Formula for Potential Energy Potential Energy of a lifted object is equal to the maximum Work that this lifted object can do by a free fall to the end down and collision there down with another object. This Work, in the event of the collision, is a pushing other object and this pushing may result the following occasions:  — the new velocities of both the objects;  — the deformation (destruction) of both the objects;  — the heat-up of both the objects, etc. Since the displacement (D), while free falling of an object, is calculated by this formula: D = a*t2/2, then, extracting the time (t), t = √(2D/a) and inserting this expression t = √(2D/a) into the Universal Formula W(E) = m*a*t, we get the New Formula for Potential Energy of a lifted object: W(E) = m*a*t = m*a*√(2D/a). For example, let’s calculate Potential Energy of the 20kg object, lifted, say, to a height of 100 meters. E(W)=20(kg)*9.8(m/s²)*√(2*100(m)/9.8(m/s²))=885.44 Энергий Thus, this 20kg object, lifted to a height of 100 meters, has such Potential Energy that, if it falls down to the end and collides there with another object, it can do a maximum Work of 885.44 Energies. It is important to understand that Spent Energy (Done Work) while lifting of an object can sometimes (in fact, almost always) not to be equal to Potential Energy of this lifted object. This new understanding also breaks the existing Law of Conservation of Energy. As already said, the New Law of Conservation of Energy is presented in the Discovery #3. The New Formula for Power. Power is the rate at which Work is done or is the rate at which Energy is consumed or given. Power (P) is: Work done per 1 second; or Energy consumed per 1 second; or Energy givven per 1 second. The new formula for Work and Energy (W(E)=F*t) automatically changes the concept of Power. And we get, Power = Force. Thus, the Power is numerically equal to the Force, i.e. to size of a pushing per 1 second. Consider the same familiar example again: the 20kg object is freely falling down. Let’s calculate the Power of the Gravitational Force:  — the Work of the gravitational Force per 1 second; or  — the increase of Kinetic Energy of the 20kg object per 1 second; or  — the decrease of Potential Energy of the 20kg object per 1 second. All these 3 things are calculated by one formula and will be equal to the same number. P=F=m*a=20(kg)*9.8(m/s²)=196 Energies per 1 second And now we have a very important conclusion: the Power of the gravitational Force is unchanged throughout the time of the free falling of the 20kg object and is always equal to 196 Energies per 1 second. The erroneousness of the Joule In this section I proved the erroneousness of using the Joule as the unit of measure for Work and Energy and the erroneousness of the “current” formula for Work and Energy. First of all, it is important to understand the meaning of the word “inertia”. “Inertia” means “inaction” (or “inactivity”). According to Newton’s 1st law, if the object has already received some velocity, then this object can move with this velocity by inertia, i.e. even in the absence of any additional pushing (i.e., with total inactivity of all Forces). Consider the same familiar example: the 20kg object is freely falling down. All that the gravitational Force (F) can do is to push this 20kg object down. The Work of the gravitational Force is precisely the pushing. As a result of this pushing the 20kg object evenly increases the velocity (uniformly accelerated). All that the gravitational Force (F) can do is to push this 20kg object down. The Work of the gravitational Force is precisely the pushing. As a result of this pushing the 20kg object evenly increases the velocity (uniformly accelerated). Thus, by creating a formula for the calculation of the Work of any pushing Force (including the gravitational Force), we (people) are trying to calculate exactly the size of the done work (the size of the done pushing). But the “current” formula for Work and Energy is erroneous, since it relates the Work (W) of the gravitational Force (F) to the displacement of the object (D). And as a result, using this erroneous formula, it turns out that if the object falls freely, the Work (W) of the gravitational Force (F) increases every second, because the object moves an increasing distance (D) each subsequent second. And this is the error of this formula. The fact is that through a free fall, the movement of the 20kg object, during every second, starting from the second second, is the result of two movements: uniformly accelerated motion and uniform motion due to inertia. Consider the Work (Pushing) of the gravitational Force in detail. So, by the beginning of the 1st second of the fall, the 20kg object has not yet accumulated the velocity v (= 0 m/s). Therefore, during the 1st second of the fall the object moves the Summary Distance D (= 4.9 m) for only one reason: - the object moves the Non-Inertial Displacement D1 (= 4.9 m) as a result of the Work (Pushing) of the gravitational Force, and - the object does not move the Inertial Displacement (D2) (the distance that is due to inertia), because the object has not yet accumulated any velocity (v) by the beginning of this second. D = D1 + D2 = 4.9m + 0m = 4.9m By the beginning of the 2nd second of the fall, the 20kg object had already accumulated the velocity v (= 9.8 m/s). Therefore, during the 2nd second of the fall, this object moves the Summary Displacement D (= 14.7 m) already for two reasons: - the object moves the Non-Inertial Displacement D1 (= 4.9 m) as a result of the Work (Pushing) of the gravitational Force; and - the object moves the Inertial Displacement D2 (= 9.8 m) as a result of movement by inertia with a velocity v (= 9.8 m/s), i.e. without the Work (Pushing) of the gravitational Force. D = D1 + D2 = 4.9m + 9.8m = 14.7m By the beginning of the 3rd second of the fall, the 20kg object had already accumulated the velocity v (= 19.6 m/s). Therefore, during the 3rd second of the fall, this object moves the Summary Displacement D (= 24.5 m) also for two reasons: - the object moves the Non-Inertial Displacement D1 (= 4.9 m) as a result of the Work (Pushing) of the gravitational Force; and - the object moves the Inertial Displacement D2 (= 19.6 m) as a result of movement by inertia with a velocity v (= 19.6 m/s), i.e. without the Work (Pushing) of the gravitational Force. D = D1 + D2 = 4.9m + 19.6m = 24.5m And so on. Thus, for this 20kg object in order to move each second the Inertial Displacement (D2), there is no need for Work (Pushing) of the gravitational Force, because the object moves by inertia, i.e. without the Work (Pushing) of the gravitational Force! It is very important to understand that in this “old” formula for Work and Energy, in the calculation of Work and Energy, the Summary Displacement (D) was erroneously taken into account, in which 2 completely different displacements were summarized: - the Non-Inertial Displacement (D1), which the object moves as a result of the Work (Pushing) of the gravitational Force; and - the Inertial Displacement (D2), which the object moves as a result of motion by inertia. We may not “give credit” to the gravitational Force for the Inertial Displacement (D2), in other words, Inertial Displacement (D2) must not increase the Work (W) of the Force (F). But now, according to this “current” formula, the Inertial Displacement (D2) erroneously increases the Work (W) of the Force (F) and, accordingly, erroneously increases the Energy (E) expended on this Work. Thus, the using of Inertial Displacement (D2) in the calculation of the Work and the Energy (that is, the calculation of the Joule) is absolutely erroneous! So, I mathematically proved the erroneousness of using the Joule as the unit of measure for Work and Energy and the erroneousness of the “current” formula for Work and Energy. This discovery has tremendous significance. Since the concept of Energy is basic in physics and since other physical concepts are based on it, even this one discovery changes all modern physics (both quantum and classical) and makes all the existing physics textbooks irrelevant and obsolete! The aftermaths of the Discovery#1 The Discovery#1 starts the new Physics.

I repeat myself cos you DO NOT GET the elenentary things. Just get it. Read it again. Read it again SLOWLY. First understand the mistake in the current formula for Work and Energy!!!!!!!!!!!!!!!!!!!!!!!!!!!!! First get the Karlson's example and about the gravitational Force (about 2 identical processes during the 2nd second ). Read the text again, again and again

Do not take on yourself too much First get the Karlson's example (about 2 identical processes during the 2nd second ). First get the fact that The 2nd second of the movement of the stone (under the action of the pushing Carlson) and the 2nd second of free fall of any stone (under the action of the pushing gravitational force) are completely identical processes. These two processes are completely identical, because in both cases: 1) at the beginning of the 2nd second the stone already has the velocity of 9.8m/s to move at this velocity by inertia (during the 2nd second); 2) as a result of pushing the outer Force (Carlson in the first case and the gravitational Force in the second case), the velocity of the stone increases by 9.8m/s (during the 2nd second); 3) the total dispacement (during the 2nd second) is: 14.7m = 9.8m (inertial displacement) + 4.9m (non-inertial displacement). Now science ERRONEOUSLY THINKS that the Energy of the gravitational Force (Karlsson’s Energy) is spent on this inertial displacement of the stone (9.8m) during the 2nd second. So NOW do you understand that the gravitational Force does not do Work moving the stone trough the 9.8m dispalcement (during the 2nd second)?

The answer will be in the Discovery#3 (about The New Law of Conservation of Energy.) Discovery #3: (temporally hidden now). If this new knowledge (that Energy of the table is spent without any movement of 20kg of the object) confuses you, just think about how many small particles are , for example, inside the iron table.These particles are pushing each other to stay together in form of the table. These pushings are Work, and, of course, Energy is wasted on this Work. Think about how much Energy is spent on these pushings every second, in order the table simply remains in the form of a firm table. And, of course, extra Energy is spent on these pushings when there is the 20kg object on the table’s surface to prevent the particles from dispersing apart from each other and to prevent this 20kg object from going downwards (this is what the 20kg object is trying to do every moment).

Did you read the 3rd link??? But I think it will be too difficult for you now. First please answer the following question. (about -- How about solving: How far does an object drop under gravity in 2 seconds?) Are you smart enough to read the text and the table So your question is How about solving: How far does an object drop under gravity in 2 seconds? The answer is 4.9m+14.7m= 19.6m Once again 4.9m (the displacement during 1st second) + 14.7m (the displacement during 2nd second) = 19.6m in 2 seconds Capishe? The answer will be in the Discovery#3 (about The New Law of Conservation of Energy.) Discovery #3: (temporally hidden now). If this new knowledge (that Energy of the table is spent without any movement of 20kg of the object) confuses you, just think about how many small particles are , for example, inside the iron table.These particles are pushing each other to stay together in form of the table. These pushings are Work, and, of course, Energy is wasted on this Work. Think about how much Energy is spent on these pushings every second, in order the table simply remains in the form of a firm table. And, of course, extra Energy is spent on these pushings when there is the 20kg object on the table’s surface to prevent the particles from dispersing apart from each other and to prevent this 20kg object from going downwards (this is what the 20kg object is trying to do every moment).

Your questions were No. That's not how this works. You answer my questions. You defend your claims.You explain the situations where you think standard physics fails, and/or how your solution is required to come up with the correct answer.I don't care about Karlson's energy. I just want to solve the problem. Solve the problem with your method. Explain why standard physics doesn't give you the right answer.Answer the questions, as you promised to do. Don't just repeat yourself. I did answer you to read again (and carefully) the the Karlsons example. You repeat, "I don't care about Karlson's energy. I just want to solve the problem. " THE KARLSON'S EXAMPLE IS NEEDED TO UNDERSTAND ABOUT THE GRAVITATIONAL FORCE So do you understand that the gravitational Force does not do Work moving the stone trough the 9.8m dispalcement (during the 2nd second)? Because by the beginning of the 1st second of the fall, the 20kg object has not yet accumulated the velocity v (= 0 m/s). Therefore, during the 1st second of the fall the object moves a total distance D (= 4.9 m) for only one reason: - the object moves the displacemen D1 (= 4.9 m) as a result of the Work (Pushing) of the gravitational Force, and - the object does not move the displacemen (D2) (the distance that is due to inertia), because the object has not yet accumulated any velocity (v). D = D1 + D2 = 4.9m + 0m = 4.9m By the beginning of the 2nd second of the fall, the 20kg object had already accumulated the velocity v (= 9.8 m/s). Therefore, during the 2nd second of the fall, this object moves the total displacement D (= 14.7 m) already for two reasons: - the object moves the displacement D1 (= 4.9 m) as a result of the Work (Pushing) of the gravitational Force; and - the object moves the displacement D2 (= 9.8 m) as a result of movement by inertia with a velocity v (= 9.8 m/s), i.e. without the Work (Pushing) of the gravitational Force. D = D1 + D2 = 4.9m + 9.8m = 14.7m So do you understand that the gravitational Force does not do Work moving the stone trough the 9.8m dispalcement (during the 2nd second)?

First of all understand that the current formula for Work and Energy is is erroneous. Do it through the Karlson's example. And then the question is So do you agree with the fact that Karlson’s Energy will not be spent on this displacement of the stone (9.8m), during the 2nd second? Because this displacement (9.8m, during the 2nd second!) would be even without Karlson. THE KARLSON'S EXAMPLE IS NEEDED TO UNDERSTAND ABOUT THE GRAVITATIONAL FORCE So do you understand that the gravitational does not do Work moving the stone trough the 9.8m dispalcement (during the 2nd second)? Because by the beginning of the 1st second of the fall, the 20kg object has not yet accumulated the velocity v (= 0 m/s). Therefore, during the 1st second of the fall the object moves a total distance D (= 4.9 m) for only one reason: - the object moves the displacemen D1 (= 4.9 m) as a result of the Work (Pushing) of the gravitational Force, and - the object does not move the displacemen (D2) (the distance that is due to inertia), because the object has not yet accumulated any velocity (v). D = D1 + D2 = 4.9m + 0m = 4.9m By the beginning of the 2nd second of the fall, the 20kg object had already accumulated the velocity v (= 9.8 m/s). Therefore, during the 2nd second of the fall, this object moves the total displacement D (= 14.7 m) already for two reasons: - the object moves the displacement D1 (= 4.9 m) as a result of the Work (Pushing) of the gravitational Force; and - the object moves the displacement D2 (= 9.8 m) as a result of movement by inertia with a velocity v (= 9.8 m/s), i.e. without the Work (Pushing) of the gravitational Force. D = D1 + D2 = 4.9m + 9.8m = 14.7m So do you understand that the gravitational does not do Work moving the stone trough the 9.8m dispalcement (during the 2nd second)?

Let's take the fact the gravitational force is constant when free-falling (since the the heights are too small). Ask your teacher about it. The Karlson's example is needed to understand the point. It's important to understand that The 2nd second of the movement of the stone (under the action of the pushing Carlson) and the 2nd second of free fall of any stone (under the action of the pushing gravitational force) are completely identical processes. These two processes are completely identical, because in both cases: 1) at the beginning of the 2nd second the stone already has the velocity of 9.8m/s to move at this velocity by inertia (during the 2nd second); 2) as a result of pushing the outer Force (Carlson in the first case and the gravitational Force in the second case), the velocity of the stone increases by 9.8m/s (during the 2nd second); 3) the total dispacement (during the 2nd second) is: 14.7m = 9.8m (inertial displacement) + 4.9m (non-inertial displacement). Now science ERRONEOUSLY THINKS that the Energy of the gravitational Force (Karlsson’s Energy) is spent on this inertial displacement of the stone (9.8m) during the 2nd second ! So do you agree with the fact that Karlson’s Energy will not be spent on this displacement of the stone (9.8m), during the 2nd second? Because this displacement (9.8m, during the 2nd second!) would be even without Karlson.

NOOOOOOOO All the exapmles about free-falling go from the assumption that the gravitational Force is constant!!!!! (the heights are too small) If the object has the mass of 20kg then the gravitational force (during the fall) is constant = 20*9.8=196 Newton !!! In other words, it can be said that the size of the pushing per any one second of falling is 196 Newtons. NOOOOOOO please read previous answer ( the gravitational force DOES NOT increase as you get closer to the ground) because the heights are too small.

The Karlson's example is needed to understand this point. It's important to understand that The 2nd second of the movement of the stone (under the action of the pushing Carlson) and the 2nd second of free fall of any stone (under the action of the pushing gravitational force) are completely identical processes. These two processes are completely identical, because in both cases: 1) at the beginning of the 2nd second the stone already has the velocity of 9.8m/s to move at this velocity by inertia (during the 2nd second); 2) as a result of pushing the outer Force (Carlson in the first case and the gravitational Force in the second case), the velocity of the stone increases by 9.8m/s (during the 2nd second); 3) the total dispacement (during the 2nd second) is: 14.7m = 9.8m (inertial displacement) + 4.9m (non-inertial displacement). Now science ERRONEOUSLY THINKS that the Energy of the gravitational Force (Karlsson’s Energy) is spent on this inertial displacement of the stone (9.8m) during the 2nd second. So do you agree with the fact that Karlson’s Energy will not be spent on this displacement of the stone (9.8m), during the 2nd second? Because this displacement (9.8m, during the 2nd second!) would be even without Karlson.

NOOOOOOOO the force does not to increase each step !!!!! Any second of fall the force (that is the size of the pushing per any second) is constant!!!!!

Please read again the 2 texts (about Karlson and the proof of the erroneousness of using the Joule as the unit of measure for Work and Energy and the erroneousness of the “current” formula for Work and Energy. ) Only repeated reading can help. The question is So do you agree with the fact that Karlson’s Energy will not be spent on this displacement of the stone (9.8m), during the 2nd second?Because this displacement (9.8m, during the 2nd second) would be even without Karlson. Keep in your mind the fact that if there is the pushing Force that provides the 9.8 (m/s^2) acceleration, then ALWAYS the displacement from this pushing is 4.9m (ONLY.) Just think about the 1st second of fall where we also have the pushing Force that provides the 9.8 (m/s^2) acceleration BUT the displacement is 4.9m. 4.9m !Understand it? First of all, it is important to understand the meaning of the word “inertia”.“Inertia” means “inaction” (or “inactivity”). When the stone is moving through this displacement (9.8m, during 2nd second) by inertia, then no Energy (of Karlson) is being spent on this displacement. The stone moves this displacement (9.8m, during 2nd second) with total inactivity of all Forces -- nothing pushes the stone to move this displacement (9.8m, during 2nd second) .

Thank you Keep in your mind the fact that if there is the pushing Force that provides the 9.8 (m/s^2) acceleration, then ALWAYS the displacement from this pushing is 4.9m (ONLY.) Just think about the 1st second of fall where we also have the pushing Force that provides the 9.8 (m/s^2) acceleration BUT the displacement is 4.9m. 4.9m ! Understand it? Thank you. First of all, it is important to understand the meaning of the word “inertia”.“Inertia” means “inaction” (or “inactivity”). When the stone is moving through this displacement (9.8m, during 2nd second) by inertia, then no Energy (of Karlson) is being spent on this displacement. The stone moves this displacement (9.8m, during 2nd second) with total inactivity of all Forces -- nothing pushes the stone to move this displacement (9.8m, during 2nd second) .

So do you agree with the fact that Karlson’s Energy will not be spent on this displacement of the stone (9.8m), during the 2nd second?Because this displacement (9.8m, during the 2nd second) would be even without Karlson.