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Cerrena [4.2K]
2 years ago
15

What is the order of magnitude of the distance of Sun to nearest star in meters?

Physics
2 answers:
Brilliant_brown [7]2 years ago
6 0

Answer:

Mercury, 46,001,272 km from the sun at the nearest point.

Explanation:

neonofarm [45]2 years ago
4 0

Answer:

Approximating the Milky Way as a disk and using the density in the solar neighborhood, there are about 100 billion stars in the Milky Way.

Explanation:

Since we are making an order of magnitude estimate, we will make a series of simplifying assumptions to get an answer that is roughly right.

Let's model the Milky Way galaxy as a disk.

The volume of a disk is:

V

=

π

⋅

r

2

⋅

h

Plugging in our numbers (and assuming that

π

≈

3

)

V

=

π

⋅

(

10

21

m

)

2

⋅

(

10

19

m

)

V

=

3

×

10

61

m

3

Is the approximate volume of the Milky Way.

Now, all we need to do is find how many stars per cubic meter (

ρ

) are in the Milky Way and we can find the total number of stars.

Let's look at the neighborhood around the Sun. We know that in a sphere with a radius of

4

×

10

16

m there is exactly one star (the Sun), after that you hit other stars. We can use that to estimate a rough density for the Milky Way.

ρ

=

n

V

Using the volume of a sphere

V

=

4

3

π

r

3

ρ

=

1

4

3

π

(

4

×

10

16

m

)

3

ρ

=

1

256

10

−

48

stars /

m

3

Going back to the density equation:

ρ

=

n

V

n

=

ρ

V

Plugging in the density of the solar neighborhood and the volume of the Milky Way:

n

=

(

1

256

10

−

48

m

−

3

)

⋅

(

3

×

10

61

m

3

)

n

=

3

256

10

13

n

=

1

×

10

11

stars (or 100 billion stars)

Is this reasonable? Other estimates say that there are are 100-400 billion stars in the Milky Way. This is exactly what we found.

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Answer:

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7 0
3 years ago
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. If a pendulum-driven clock gains 5.00 s/day, what fractional change in pendulum length must be made for it to keep perfect tim
inn [45]

Answer:

The appropriate response will be "Length must be increased by 0.012%".

Explanation:

The given values is:

ΔT = 5 s/day

Now,

⇒ \frac{\Delta T}{T} =\frac{5}{24\times 60\times 60}

On multiplying both sides by  "100", we get

⇒ \frac{\Delta T}{T}\times 100 =\frac{500}{24\times 60\times 60}

⇒ \frac{\Delta T}{T}\times 100=0.005787 (%)

∵  T=2\pi\sqrt{\frac{l}{g} }

On substituting the values, we get

⇒ \frac{\Delta T}{T}% = \frac{1}{2}\times \frac{\Delta l}{l}%

On applying cross multiplication, we get

⇒ \frac{\Delta l}{l}% = 2\times \frac{\Delta T}{T}%

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6 0
3 years ago
Earth is about 150 million kilometers from the Sun, and the apparent brightness of the Sun in our sky is about 1300 watts/m2. Us
nalin [4]

Answer:

13 W/m^2

Explanation:

The apparent brightness follows an inverse square law, therefore we can write:

I \propto \frac{1}{r^2}

where I is the apparent brightness and r is the distance from the Sun.

We can also rewrite the law as

\frac{I_2}{I_1}=\frac{r_1^2}{r_2^2} (1)

where in this problem, we have:

I_1 = 1300 W/m^2 apparent brightness at a distance r_1, where

r_1 = 150 million km

We want to estimate the apparent brightness at r_2, where r_2 is ten times r_1, so

r_2 = 10 r_1

Re-arranging eq.(1), we find I_2:

I_2 = \frac{r_1^2}{r_2^2}I_1 = \frac{r_1^2}{(10r_1)^2}(1300)=\frac{1}{100}(1300)=13 W/m^2

5 0
3 years ago
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olasank [31]
Ionic bonds with electrostatic attractions
4 0
3 years ago
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A pendulum has 711 J of potential energy at the highest point of its swing. How much kinetic energy will it have at the bottom o
Anika [276]
According to law of conservation of energy, 
<span>Energy can neither be constructed nor be destroyed but can be transformed from one form to another.
</span>
<span>At the highest point of the pendulum(point b), pendulum is associated with potential energy only and no kinetic energy.
</span><span>Therefore total energy at point b = potential energy = 711 J.... i
</span>
<span>At the bottom most point(point a), pendulum is associated only with kinetic energy and no potential energy.
</span>Therefore total energy at point a = kinetic energy ---- ii
<span>From i and ii,
</span>Kinetic energy = potential energy = 711 J.(Conserving energy)

Hence kinetic energy at the bottom most point is 711 J.
Hope this helps!!

7 0
2 years ago
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