Universe
11th Edition
ISBN: 9781319039448
Author: Robert Geller, Roger Freedman, William J. Kaufmann
Publisher: W. H. Freeman
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Chapter 16, Problem 8Q
(a)
To determine
The amount of energy released, if a carbon
(b)
To determine
The amount of energy released, if mass of
(c)
To determine
The amount of energy released, if a planet of mass
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The sun produces energy by nuclear fusion reactions, in which matter is converted into energy. By measuring the amount of energy we receive from the sun, we know that it is producing energy at a rate of 3.8 x 1026 W. (a) How many kilograms of matter does the sun lose each second? Approximately how many tons of matter is this (1 ton = 2000 lb)? (b) At this rate, how long would it take the sun to use up all its mass?
A 4.6 kg rocket is launched directly upward from Earth at 9.00 km/s (rE = 6.38 x 10^ 6m, mE = 5.98 x 10^24 kg, G =6.67 x 10^-11N. m2 /kg2)a) What altitude above Earth's surface does the rocket reach?
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Chapter 16 Solutions
Universe
Ch. 16 - Prob. 1CCCh. 16 - Prob. 2CCCh. 16 - Prob. 3CCCh. 16 - Prob. 4CCCh. 16 - Prob. 5CCCh. 16 - Prob. 6CCCh. 16 - Prob. 7CCCh. 16 - Prob. 8CCCh. 16 - Prob. 9CCCh. 16 - Prob. 10CC
Ch. 16 - Prob. 11CCCh. 16 - Prob. 12CCCh. 16 - Prob. 13CCCh. 16 - Prob. 14CCCh. 16 - Prob. 15CCCh. 16 - Prob. 16CCCh. 16 - Prob. 17CCCh. 16 - Prob. 18CCCh. 16 - Prob. 19CCCh. 16 - Prob. 1CLCCh. 16 - Prob. 2CLCCh. 16 - Prob. 1QCh. 16 - Prob. 2QCh. 16 - Prob. 3QCh. 16 - Prob. 4QCh. 16 - Prob. 5QCh. 16 - Prob. 6QCh. 16 - Prob. 7QCh. 16 - Prob. 8QCh. 16 - Prob. 9QCh. 16 - Prob. 10QCh. 16 - Prob. 11QCh. 16 - Prob. 12QCh. 16 - Prob. 13QCh. 16 - Prob. 14QCh. 16 - Prob. 15QCh. 16 - Prob. 16QCh. 16 - Prob. 17QCh. 16 - Prob. 18QCh. 16 - Prob. 19QCh. 16 - Prob. 20QCh. 16 - Prob. 21QCh. 16 - Prob. 22QCh. 16 - Prob. 23QCh. 16 - Prob. 24QCh. 16 - Prob. 25QCh. 16 - Prob. 26QCh. 16 - Prob. 27QCh. 16 - Prob. 28QCh. 16 - Prob. 29QCh. 16 - Prob. 30QCh. 16 - Prob. 31QCh. 16 - Prob. 32QCh. 16 - Prob. 33QCh. 16 - Prob. 34QCh. 16 - Prob. 35QCh. 16 - Prob. 36QCh. 16 - Prob. 37QCh. 16 - Prob. 38QCh. 16 - Prob. 39QCh. 16 - Prob. 40QCh. 16 - Prob. 41QCh. 16 - Prob. 42QCh. 16 - Prob. 43QCh. 16 - Prob. 44QCh. 16 - Prob. 45QCh. 16 - Prob. 46QCh. 16 - Prob. 47QCh. 16 - Prob. 48QCh. 16 - Prob. 50QCh. 16 - Prob. 51QCh. 16 - Prob. 52QCh. 16 - Prob. 53QCh. 16 - Prob. 54QCh. 16 - Prob. 55QCh. 16 - Prob. 56QCh. 16 - Prob. 57QCh. 16 - Prob. 58QCh. 16 - Prob. 59QCh. 16 - Prob. 60Q
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- Since 1995, hundreds of extrasolar planets have been discovered. There is the exciting possibility that there is life on one or more of these planets. To support life similar to that on the Earth, the planet must have liquid water. For an Earth-like planet orbiting a star like the Sun, this requirement means that the planet must be within a habitable zone of 0.9 AU to 1.4 AU from the star. The semimajor axis of an extrasolar planet is inferred from its period. What range in periods corresponds to the habitable zone for an Earth-like Planet orbiting a Sun-like star?arrow_forwardIn fact, the conversion of mass to energy in the Sun is not 100% efficient. As we have seen in the text, the conversion of four hydrogen atoms to one helium atom results in the conversion of about 0.02862 times the mass of a proton to energy. How much energy in joules does one such reaction produce? (See Appendix E for the mass of the hydrogen atom, which, for all practical purposes, is the mass of a proton.)arrow_forwardWhat is the total energy possessed by a satellite of mass “m” orbiting above the earth of mass “M” and radius “R” at a height “h”?a) [(G*M*m)/2*(R+h)]b) -[(G*M*m)/2*(R+h)]c) [(G*M*m)/(R+h)]d) -[(G*M*m)/(R+h)]arrow_forward
- The carbon isotope 14C is used for carbon dating of objects. A 14C nucleus can change into a different kind of element, a neighbor on the periodic table with lower mass, by emitting a beta particle – an electron or positron – plus a neutrino or an anti-neutrino. Consider the scenario where 14C ( mass of 2.34 x 10 -26) decays by emitting an electron and anti neutrino. The electron has a mass of 9.11x 10-31 kg and a speed of 1.0 x107 m/s. While the anti neutrino has a momentum of 4.0x10-24 kg-m/s. If the electron and anti neutrino are emitted at right angles from each other, calculate the recoil speed of the nucleus.arrow_forwardSuppose 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.arrow_forwardThe Sun's mass is1.989 ×10^8 and it radiates at a rate of 3.827×10^23 kW. a) From this data, assuming it converts all its mass into energy, what is the estimate the lifetime of the Sun? b) Theoretical calculations predict the Sun's lifetime (in its current stage) to be about 5 billion years. During that time, what percentage of its mass will it lose?arrow_forward
- A unique star produces energy by nuclear fusion reactions, in which matter is converted into energy. By measuring the amount of energy we receive from that star, we know that it is producing energy at a rate of 4.8 x 10 26 How many kilograms of matter does this star lose each second?arrow_forwardGravitational Equilibrium: A stable orbit of an object of mass m around another object of mass M requires that the energy of motion or “kinetic” energy of the object be a certain proportion of its gravitational binding energy or “potential” energy. The kinetic energy, KE, is defined KE=1/2mv^2 where m is the mass of the object and v is its speed. The gravitational potential energy is defined as U=-(GMm)/(r) where G is the gravitational constant of Newton’s so-called Universal Law of Gravitation, M and m are the masses of the interacting objects and r is the radius of the orbit. The total energy, E, of the object of mass m is the sum of its kinetic and potential energies, E = KE + U. Note the minus sign in the definition of potential energy. This can be interpreted as the amount of energy an object has to overcome in order to escape the gravitational influence of another object. (Imagine rolling a marble up the surface of the inside of a bowl - see below. The well…arrow_forwardThe rest energy E of an object with rest mass m is given by Albert Einstein’s famous equation E = mc2, where c is the speed of light in vacuum. Find E for an electron for which (to three significant figures) m = 9.11 * 10^-31 kg. The SI unit for E is the joule (J); 1 J = 1 kg m2/s2. c = 2.99792458 * 10^8 m/s.arrow_forward
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