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alekssr [168]
3 years ago
7

Neutron stars consist only of neutrons and have unbelievably high densities. a typical mass and radius for a neutron star might

be 9.9 x 1028 kg and 1.9 x 103 m. (a) find the density of such a star. (b) if a dime (v = 2.0 x 10-7 m3) were made from this material, how much would it weight (in pounds)?
Physics
1 answer:
Tanya [424]3 years ago
4 0
<span>Density is 3.4x10^18 kg/m^3 Dime weighs 1.5x10^12 pounds The definition of density is simply mass per volume. So let's divide the mass of the neutron star by its volume. First, we need to determine the volume. Assuming the neutron star is a sphere, the volume will be 4/3 pi r^3, so 4/3 pi 1.9x10^3 = 4/3 pi 6.859x10^3 m^3 = 2.873x10^10 m^3 Now divide the mass by the volume 9.9x10^28 kg / 2.873x10^10 m^3 = 3.44588x10^18 kg/m^3 Since we only have 2 significant digits in our data, round to 2 significant digits, giving 3.4x10^18 kg/m^3 Now to figure out how much the dime weighs, just multiply by the volume of the dime. 3.4x10^18 kg/m^3 * 2.0x10^-7 m^3 = 6.8x10^11 kg And to convert from kg to lbs, multiply by 2.20462, so 6.8x10^11 kg * 2.20462 lb/kg = 1.5x10^12 lb</span>
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sveta [45]

Answer:

a) Average\ Speed=28\ km/h\\b) Average\ Velocity= 20\ km/h

Explanation:

We\ are\ given\ that,\\Velocity\ of\ the\ car\ Eastwards=60\ km/h\\Time\ taken\ by\ the\ car\ Eastwards=2\ h\\Velocity\ of\ the\ car\ Northwards=20\ km/h\\Time\ taken\ by\ the\ car\ Northwards=8\ h\\Hence,\\As\ we\ know\ that,\\Speed=\frac{Distance}{Time}\\Distance= Speed* Time\\Now,\ lets\ find\ the\ distance\ covered\ by\ the\ car\ in\ both\ the\ cases.

Hence,\\Distance\ Covered\ During\ its\ Eastward\ Journey=60*2=120\ km\\Distance\ Covered\ During\ its\ Northward\ Journey=20*8=160\ km\\Now,\\As\ we\ know\ that\ Average\ Speed=\frac{Total\ Distance}{Total\ Time} \\Here,\\Total\ distance\ of\ the\ car=Distance\ Covered\ During\ its\ Northward\ Journey+Distance\ Covered\ During\ its\ Eastward\ Journey\\Hence,\\Total\ distance\ of\ the\ car=120+160=280\ km\\Total\ time\ taken\ by\ the\ car=8+2=10\ hours\\Hence,\\Average\ Speed\ Of\ the\ Car\ throughout\ its\ journey=\frac{280}{10}=28\ km/h

Now,\\For\ Average\ Velocity\ we\ need\ to\ consider\ displacement\ as:\\Average\ Velocity\ =\frac{Total\ Displacement}{Total\ Time} \\Now,\\As\ we\ already\ know\ that\ displacement\ is\ the\ shortest\ distance\\ from\ the\ initial\ to\ the\ final\ point.\\We\ observe\ that, \\The\ car\ forms\ a\ right\ triangle\ during\ its\ complete\ journey.\\Hence,\\As\ we\ already\ know\ that,\\Distance\ travelled\ Eastwards= 120\ km\\Distance\ travelled\ Northwards= 160\ km\\Hence,\\We\ may\ apply\ Pythagoras\ Property\ to\ find\ the\ net\ displacement.\\Hence,\\a^2+b^2=c^2\\120^2+160^2=c^2\\14400+25600=c^2\\40000=c^2\\c=\sqrt{40000}\\c=200\\Hence,\\Total\ displacement=200\ km\\Total\ Time\ taken=2+8=10\ hours\\Hence,\\Average\ Velocity\ Of\ the\ Car=\frac{200}{10}=20\ km/h

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3 years ago
A wire is oriented along the x-axis. It is connected to two batteries, and a conventional current of 2.6 A runs through the wire
lana66690 [7]

Answer:

the magnitude of the magnetic force on the wire is 0.2298 N

Explanation:

Given the data in the question;

we know that, the magnitude of magnetic force is given as;

|F_{mg}^> | = I(B^> × L^> )

given that

I = 2.6 A

B^> = 0.17

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|F_{mg}^> | = 2.6( 0.17i" × 0.52j" )

|F_{mg}^> | = 0.2298 N

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