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In Exercise 12.4 of your book,University Physics 15th edition (see End of the Chapter 12section), what is the answer for sub-item (b) if theradius of the neutron st...

Question

In Exercise 12.4 of your book,University Physics 15th edition (see End of the Chapter 12section), what is the answer for sub-item (b) if theradius of the neutron star is 73.09 km? (express your answer in theproper SI unit and without scientificnotation)

In Exercise 12.4 of your book, University Physics 15th edition (see End of the Chapter 12 section), what is the answer for sub-item (b) if the radius of the neutron star is 73.09 km? (express your answer in the proper SI unit and without scientific notation)



Answers

A neutron star is a nucleus composed entirely of neutrons and has a radius of about $10 \mathrm{km}$. Such a star forms when a larger star cools down and collapses under the influence of its own gravitational field. The compression continues until the protons and electrons merge to form neutrons. (a) Assuming that Equation 13.1 is valid, estimate the mass number$A$ and mass $M$ of a neutron star of radius $R=10 \mathrm{km}$(b) Find the acceleration due to gravity at the surface of such a star. (c) Find the rotational kinetic energy of the star if it rotates about its axis 30 times/s. (Assume that the star is a sphere of uniform density.)

Right. So for this question, we're looking at a hypothetical situation. When, What it What happens if you son transformed into a neutron star? What would be the radius right north, This off the sun. So we know that Ah, the mass of the sun. It's just 1.99 times stand about 30 kilograms. And if it were a neutron star, it would just be made up off new clearance. Right? You can't will be stacked upon 11 on top off the other. And the density off the sun would be equals two D tense city off a nuclear. Right. So for the density often, Yukun, we need to take the mess off any new corn divided by its volume, which is a previous question. And ah, you can find that for any arbitrary new cone with any you can number a, it will cancel well, and it would give you a constant value. And this is to country time stand toe power off 17 kilograms per meter cubed. Now, using this density, we cannot quit to find what is the radius, right? We don't know what's the radius. So if it were to have a radius off our then it's fall. You would be faulted pi r cubed. This would be its volume. And to get this mess, we need to multiply with the density and the density just given already. Over here, right? Just you You should be the density and this we can create to the mess off the sun. So the only unknown variable here, which is all radius we can find, Right So are is this sun for the by for two by rule take to the power off one thing Yeah, Put a talk calculator. Our final results should be about 30. Give me this.

As we all know, that volume is equal to four by three by our cube which is equal to mass by density. So I can write The value of R is equal to T d m bye four. By rule hold to the power one by three on solving it further I can write the value of a ridge three into one into 10 to the power 30 kg by food by multiplication 2.3 into 10 to the power 17 Hold to the power one by three On solving it further I get the value of our ed Turn to the powerful meter which is equal to then to the powerful by 10 to the power three kilometer I just change the unit So R is equal to 10 kilometers is our answer

Question 80 axe are states that neutron sirs are so named because they're composed of neutrons. Very density, the same as that of nucleus. Referring to example, 32-4 for the nuclear density. Find the radius of a neutron star whose masses half that of the sun, so make sense, huh? Um, neutron stars are composers. Nutrients, as one might expect. Um, from that example, 32 4 were given, um, this expression for density, Which is true for saying for this neutron star so we can use that example here the masses half the sun mass, which I calculated to be 0.995 times in the 30 kilograms. And so we won't find the radius of this object of this neutron. Sir, you know, for mass and density, so that should give you a hint that we need the relationship. Long's he density is just mass over volume. And hey, volume has a radius term in it. So we can rearrange this to solve for William, simply be mass over density. If we consider the sun or any start really to be a sphere, then obviously the volume of our sphere is four pi r cubed over three mass over density such that if we continue on over here, the radius of a star can be represented as three times the mass over four pi times the density all to the power of 1/3. Still, since we're given Mass and were given density, we can plug this in Jordan equation. Um, yeah, everybody's given. So we find that our resulting radius of our star to be 1.1 times 10 to the four meters or we can represent this. I guess we have to see figures you started based on our density, so we could just call it a 10 kilometer radius star. So having half the size of this half the mass of the earth Sorry, half the mass of the sun. This star has only a 10 kilometer radius. That is absurd. That is such a high density of mass bad, crazy

So there's several steps involved in this problem. First, we recognize that and neutron stars made up of mostly neutrons. If we assume that a neutron star has the same density as that of a neutron itself, then the first thing we need to do is calculate the density of a neutron. We can do that by before getting the density calculated. The volume of a neutron. The volume of a neutron is going to be equal to four thirds pi r cubed, where R is given to us at 1.0 times 10 to the negative 13 centimeters. This then gives us 4.19 times 10 to the negative, 39 centimeters cubed. How do we get density? We need a mass. We can go to our table for one, and we see that the mass of the neutron is 1.67493 10 of the negative 27 kg. If we convert that into grams, it's 1.67493 times 10 to the negative 24 g density, then is mass over volume, so we'll take the mass of a single neutron in grams divided by the volume in centimeters cubed, and we get 4.0 times 10 to the 14 g per centimeter cubed. Now, assuming that the entire star has the density of a neutron, the entire neutron star, then if we have a pebble size of a neutron star, with the pebble having a radius of 0.1 millimeters. In order to calculate the mass of that, the first thing that we need to do is calculate the volume of that pebble sized neutron star with a radius of 0.1 millimeters. Because our density has units of centimeters cubed, we need to convert our millimeters into centimeters. 0.1 millimeter is 0.1 centimeters. We can then take 0.1 centimeters, plug it into our four thirds pi r cubed equation, and we'll get a volume of that pebble sized neutron star 4.19 times, 10 to the negative, six centimeters cubed we can then use density is a conversion factor and multiply it by the volume we calculated, and we'll get grams of 1.67 times 10 to the 9 g. So that's ah pretty heavy Pebble


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