The point of this question is to compare rest energy and kinetic energy at high speeds.An alpha particle (a helium nucleus) is moving at a speed of 0.9994 times the speed of light. Its mass is (6.40 x 10-27 kg).(a) What is its rest energy? 5.75e-10(b) Is it okay to calculate its kinetic energy using the formula mv2?O No, because this formula isn't valid for speeds near the speed of light.O Yes, because v << c.O Yes, because the formula mv² for kinetic energy is always correct.(c) What is its kinetic energy?(d) Which is true?O The kinetic energy is approximately equal to the rest energy.O The kinetic energy is much bigger than the rest energy.O The kinetic energy is much smaller than the rest energy. One mole of helium atoms has a mass of 4 grams. If a helium atom in a balloon has a kinetic energy of 1.666e-21 J, what is the speed of the helium atom? (The speed is much lower than the speed of light.)V =m/sAdditional MaterialsO eBook

Given:- mass = (6.40 x 10-27 kg) Speed = 0.9994 C where C is the speed of light Find:- a) The…

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College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Consider the equation for kinetic energy: KE = 1/2mv^2 = 1/2 * m * v^2. If I ask you to take the derivative of kinetic energy, you should ask "the derivative with respect to what?" a) Suppose mass m is constant. Compute the derivative of KE with respect to v, (d(KE)/dv). b) Who takes derivatives with respect to velocity? No one. Except you, just now. Sorry. The rate of change of energy with respect to time is more important: it is the Power. Now, consider velocity v to be a function of time, v(t). We will rewrite KE showing this time dependance: KE= 1/2 * m * v(t)^2. Show that (d(KE)/dt) = F(t)v(t). Hint: use Newton's second law, F = ma, to simplify. c) In the computation above, we assumed m was constant, and v was changing in time. Think of a physical situation in which both m and v are varying in time. d) Compute the Power when both mass and velocity are changing in time. (First rewrite KE(t) showing time dependence, then compute (d(KE)/dt).
Suppose that you have found a way to convert the rest energy of any type of matter directly to usable energy with an efficiency of 61.0%. How many liters of water would be sufficient fuel to very slowly push the Moon 1.70 mm away from the Earth? The density of water is ?water=1.00kg/liter, the Earth's mass is ?earth=5.97×1024 kg, the Moon's mass is ?moon=7.36×1022 kg, and the separation of the Earth and Moon is ?E,M=3.84×108 m.
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The escape velocity from a massive object is the speed needed to reach an infinite distance from it and have just slowed to a stop, that is, to have just enough kinetic energy to climb out of the gravitational potential well and have none left. You can find the escape velocity by equating the total kinetic and gravitational potential energy to zero E=12mv2esc−GmM/r=0E=12mvesc2−GmM/r=0 vesc=2GM/r−−−−−−√vesc=2GM/r where GG is Newton's constant of gravitation, MM is the mass of the object from which the escape is happening, and rr is its radius. This is physics you have seen in the first part of the course, and you should be able to use it to find an escape velocity from any planet or satellite. For the Earth, for example the escape velocity is about 11.2 km/s, and for the Moon it is 2.38 km/s. A very important point about escape velocity: it does not depend on what is escaping. A spaceship or a molecule must have this velocity or more away from the center of the planet to be free…
Hannah the Standard Poodle has just been groomed, and without so much fur she weighs in at a lithe 21 kg. What mass of fuel would she need to convert entirely into energy in order to accelerate from rest to 1% of the speed of light? [c = 2.998 x 10^8 m/s] PLEASE sketch the situation correct answer is 1.1 g
A spaceship of mass 2.00 x 106 kg is to be accelerated to a speed of 0.730c. (a) What minimum amount of energy does this acceleration require from the spaceship's fuel, assuming perfect efficiency? 19.23e22 X Your response differs from the correct answer by more than 100%. J (b) How much fuel would it take to provide this much energy if all the rest energy of the fuel could be transformed to kinetic energy of the spaceship? 2.136e6 X Your response differs from the correct answer by more than 100%. kg

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