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g100num [7]
3 years ago
5

In about 5 billion years, at the end of its lifetime, our sun will end up as a white dwarf, having about the same mass as it doe

s now, but reduced to about 15,000 km in diameter.What will be its density at that stage?
...?
Physics
1 answer:
vitfil [10]3 years ago
8 0
It is fine to use the equation given by Plitter, since we are told that the mass is about the same as it is now, and I seriously doubt the original question wants the student to go into relativistic effects, electron degeneracy pressure and magnetic effects that govern a real white dwarf star.
There is no need to make it unnecessarily complicated, when the question is set at high school level.  The question asks, given a particular radius, and a given mass, what will the density be (which in this case will be the average density).   To answer the question, one needs to know the mass of the sun (which is about 2×1030 Kg.  One needs to convert the diameter to a radius, and then calculate the spherical volume of the white dwarf.  Then one can use the formula given above, namely density=mass/volume
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A wooden block meauring 40cm x 10cm x 5cm has a mass 850gm . find the density of wood?
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Answer:

Explanation:

Density = Mass / Volume = 850 / 40*10*5 = 0.425 g /cm^3

5 0
3 years ago
A cylindrical wire has a length of 2.80 m and a radius of 1.03 mm. It carries a current of current of 1.35 A, when a a voltage o
aleksley [76]

Answer:

0.023 Ohms

Explanation:

Given data

Length= 2.8m

radius= 1.03mm

current I= 1.35 A

voltage V= 0.032V

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V= IR

Now  R= V/I

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R= 0.032/1.35

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3 0
3 years ago
A negative point charge q1 = 25 nC is located on the y axis at y = 0 and a positive point charge q2 = 10 nC is located at y =14
sergey [27]

Answer:

 y = 0.1 m

Explanation:

The electrical power for point loads is

         V = k \sum \frac{q_i}{r_i}k Sum qi / ri

in this case

         V = k (- \frac{q_1}{r_1 } + \frac{q_2}{r_2})

indicate that V = 0

        \frac{q_1}{r_1} = \frac{q_2}{r_2}

        r₂ = \frac{q_2}{q_1} r_1

the distance r1 is

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the distance r2

         r₂ = 0.14 -y

we substitute

       

        0.14 - y = \frac{10}{25}  y

          y ( \frac{10}{25} + 1) = 0.14

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          y = 0.14 / 1.4

          y = 0.1 m

7 0
3 years ago
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