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An antimatter reactor. When a particle meets its antipar-ticle (more about this in Chapter 30 , they annihilate eachother and their mass is converted to light energy. The UnitedStates uses approximately 1.0 $\times 10^{20} \mathrm{J}$ of energy per year. (a) Ifall this energy came from a futuristic antimatter reactor, howmuch mass would be consumed yearly? (b) If this antimatterfuel had the density of Fe $\left(7.86 / \mathrm{cm}^{3}\right)$ and were stacked inbricks to form a cubical pile, how high would it be? (Beforeyou get your hopes up, antimatter reactors are a long way inthe future-if they ever will be feasible.)

52.08 $\mathrm{cm}$

Physics 101 Mechanics

Chapter 27

Relativity

Gravitation

Simon Fraser University

Hope College

University of Winnipeg

McMaster University

Lectures

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In physics, orbital motion is the motion of an object around another object, which is often a star or planet. Orbital motion is affected by the gravity of the central object, as well as by the resistance of deep space (which is negligible at the distances of most orbits in the Solar System).

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Sir Isaac Newton described the law of universal gravitation in his work "Philosophiæ Naturalis Principia Mathematica" (1687). The law states that every point mass attracts every single other point mass by a force pointing along the line intersecting both points. The force is proportional to the product of the two masses and inversely proportional to the square of the distance between them.

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So in this problem we have imaginary antimatter reactor and we know that the total energy consumed by the United States in one year is just one times 10 to the hour of 20 gels. So the first I 10 of this problem, we want to calculate how much mass of this matter. Oh, anti matter should be consumed to produce this kind of energy. So you just need to remember at the total energy off the particle is a particle. At rest should be m C square. So if we already know what should be the energy, we can say that the mess should be if divided by C square. So this is just one times tend to the power off 20 divided by the speed of light, richest three times 10 to the power of eight square. So if we calculate this, this is 16 desist warranty. So we're going to get one point 11 times 10 your power of tree kilograms. And that's the first answer that we have the answer just home. How much mess should be used to produce this amount of energy? So the second item we want to calculate what should be the height of a bio off this material. If the material has the density of the irony s O, we know that density is just mass divided by volume. So if we have the density off Eireann, that is sh that is seven point 86 times 10 to the power of three. We can calculate this in here and should be volume equals mass divided by density, which gives us 1.11 times 10. The power of three, divided by seven point. It is six times 10 to the power of three. This we can cross. So this is precisely 0.1 41 meters Q Bix. So we want to know the height off this fire, not his volume. So we can see that lense or height should be the volume in the coup Big square. So this is just, uh, zero point 1 41 into cubic square, and this should be 0.5 to 1 meters or for simplify the height of the fire that should be necessary to produce. This amount of energy is just 52.1 ST Emitters. Only this amount of mass is necessary to produce all this energy. That's all. Thanks, Roy

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