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DIA [1.3K]
3 years ago
7

Consider the 1-to10^19 scale on which the disk of the Milky Way Galaxy fits on a football field. On this scale, how far is it fr

om the sun to the alpa centauri (real distance:4.4 light-years) How big is the sun itself on this scale? Compare the sun's size on this scale to the actual size of a typical atom (about 10^-10 m in diameter).
Mathematics
2 answers:
julia-pushkina [17]3 years ago
8 0

Answer:

Sun is 0.4162708 cm away from alpha century.

Sun is 1.391016\times10^{-10}m.

Sun is 1.391016 time the size of an atom at this scale.

Step-by-step explanation:

Light year is a measure of distance. It is the distance light travels in an year.

Light year = 9.4607\times10^{12} km

So 4.4 light years = 4.4\times9.4607\times10^{12} km

                                41.62708\times10^{12} km

Lets scale this down to the level of 1\times10^{-19}

41.62708\times10^{12}\times1\times10^{-19} km

=41.62708\times10^{-7} km

Change the units to centimeters:

41.62708\times10^{-7}\times1\times10^{5} cm

=41.62708\times10^{-2} cm

=0.4162708 cm

Therefore on the new scale sun is 0.4162708 cm away from alpha century.

Diameter of the sun is 1.391016 million km

Lets change Sun's diameter to the new scale:

1.391016\times10^{6} \times10^{-19}km

=1.391016\times10^{-13}km

Lets change kilometers in to meters:

1.391016\times10^{-13}\times10^{3}m

=1.391016\times10^{-10}m

Therefore, sun is 1.391016\times10^{-10}m

and an atom is 1\times10^{-10}

Therefore the sun is 1.391016 time the size of an atom at this scale.



bezimeni [28]3 years ago
5 0

Beautiful !  What a great exercise !  I won't even check to verify
that 1:10⁻¹⁹ is an appropriate scale to map the Milky Way onto a
football field.  I'll just accept it and go from there.

(4.4 light years) x (10⁻¹⁹) = (2.5848 x 10¹³ miles) x (10⁻¹⁹)

                                       =  2.5848 x 10⁻⁶ mile

                                       =    0.164 inch .

(1,391,400 kilometers) x (10⁻¹⁹) = 1.3914 x 10⁻¹⁰ meter


Research Conclusions:

If the Milky Way shrinks to fit on a football field, then

-- Alpha Centauri is  0.164 inch from us.

-- The sun's diameter is about 1.39 times that of a Hydrogen atom. 

Playing with this same fun concept a little more:

-- The Earth is sailing around in its orbit, once a year,
    about 0.000015 millimeter from the sun. 

-- It takes light almost 27 years to travel 1 inch !

wow  !
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One ordered pair (a,b) satisfies the two equations ab^4 = 384 and a^2 b^5 = 4608. What is the value of a in this ordered pair?
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Answer: a = 3∛2

<u>Step-by-step explanation:</u>

ab⁴ = 384     --> a = 384/b⁴

Substitute a = 384/b⁴ into the second equation to solve for "b".

   a²b⁵ = 4608

\bigg(\dfrac{384}{b^4}\bigg)^2\cdot b^5=4608\\\\\\\dfrac{147,456b^5}{b^8}=4608\\\\\\\dfrac{147,456}{b^3}=4608\\\\\\\dfrac{147,456}{4608}=b^3\\\\\\32=b^3\\\\\\\sqrt[3]{32} =b\\\\\\2\sqrt[3]{4} =b

Substitute b = 2∛4 into the first equation to solve for "a".

ab⁴ = 384

a(2∛4)⁴ = 384

        a  = 384/(2∛4)⁴

        a = 24/4∛4

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Step-by-step explanation:

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Ten kids can go to the water park for $95. How many can go if there is $200?
posledela

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Pythagorean theorem If l = 12 cm and m = 35 cm, what is the length of n?
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If a radioactive isotope has a half-life of 6 hours and 12 milligrams is initially present, how much will
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Answer:0.75 milligrams will  be present in 24 hours

Step-by-step explanation:

Step 1

The formula for radioactive decay can be written as

N(t)=No (1/2)^(t/t 1/2)

where

No= The amount of the radioactive substance at time=0=milligrams

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Step 2--- Solving

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