COLLEGE PHYSICS
2nd Edition
ISBN: 9781464196393
Author: Freedman
Publisher: MAC HIGHER
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Question
Chapter 27, Problem 39QAP
To determine
The new radius of the sun when it collapses into a neutron star.
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Part A
Some atomic masses
Particle
Symbol
Mass (u)
Electron
e
0.00055
Part B
Proton
1.00728
Neutron
n
1.00866
Part C
Hydrogen
1.00783
Helium
4 Не
4.00260
Part D
Part E
Calculate the binding energy EB of the helium nucleus He.
Express your answer in megaelectron volts to three significant figures.
• View Available Hint(s)
EB =
MeV
|The Electric potential on the axis
of a uniformly charged ring of radius R, at
a distance x from its center (x < R) is
kQ
given by: V =
Select one:
True
O False
I Review I Constants
During most of its lifetime, a star maintains an equilibrium size in
which the inward force of gravity on each atom is balanced by an
outward pressure force due to the heat of the nuclear reactions in
the core. But after all the hydrogen "fuel" is consumed by nuclear
fusion, the pressure force drops and the star undergoes a
gravitational collapse until it becomes a neutron star. In a neutron
star, the electrons and protons of the atoms are squeezed
together by gravity until they fuse into neutrons. Neutron stars
spin very rapidly and emit intense pulses of radio and light waves,
one pulse per rotation. These "pulsing stars" were discovered in
the 1960s and are called pulsars.
Part A
= 2.0 x 1030 kg) and size (R
3.5 x 10° m) of our sun rotates once every 35.0 days. After undergoing gravitational collapse,
A star with the mass (M
the star forms a pulsar that is observed by astronomers to emit radio pulses every 0.100 s. By treating the neutron star as a solid sphere,…
Chapter 27 Solutions
COLLEGE PHYSICS
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- Our solar system orbits the center of the Milky Way galaxy. Assuming a circular orbit 30,000 ly in radius and an orbital speed of 250 km/s, how many years does it take for one revolution? Note than this is approximate, assuming constant speed and circular orbit, but it is representative of the time for our system and local stars to make one revolution around the galaxy.arrow_forwardUsing data from the previous problem, find the increase in rotational kinetic energy, given the core’s mass is 1.3 times that of our Sun. Where does this increase in kinetic energy come from?arrow_forward*•58 Two radioactive materials that alpha decay, 238U and 232Th, and one that beta decays, "K, are sufficiently abundant in granite to contribute significantly to the heating of Earth through the de- cay energy produced. The alpha-decay isotopes give rise to decay chains that stop when stable lead isotopes are formed. The isotope 4"K has a single beta decay. (Assume this is the only possible decay of that isotope.) Here is the information: Stable Decay Half-Life End Parent Mode (y) Point (MeV) (ppm) 238U 232Th 4.47 x 10° 206рЬ 51.7 1.41 x 1010 208Pb 42.7 13 1.28 x 10° 40Ca 1.31 4 In the table Q is the total energy released in the decay of one par- ent nucleus to the final stable end point and f is the abundance of the isotope in kilograms per kilogram of granite; ppm means parts per million. (a) Show that these materials produce energy as heat at the rate of 1.0 x 10-9 W for each kilogram of granite. (b) Assuming that there is 2.7 x 102 kg of granite in a 20-km-thick spherical shell at…arrow_forward
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