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anygoal [31]
3 years ago
8

If the solar system shrank so that the sun were located just one centimeter from Earth, about how far away could you find the ne

xt nearest star?
Physics
1 answer:
DedPeter [7]3 years ago
8 0
E                  S                                                               *

The "E" represents Earth, "S" represent Sun, and the "*" represents the nearest star(which is Proxima Centauri).

The main thing to worry about here is units, so ill label everything out.
D'e,s'(Distance between earth and sun) = .<span>00001581 light years
D'e,*'(Distance between earth and Proxima) = </span><span>4.243 light years

Now this is where it gets fun, we need to put all the light years into centimeters.(theres alot)
In one light year, there are </span>9.461 * 10^17 centimeters.(the * in this case means multiplication) or 946,100,000,000,000,000 centimeters.

To convert we multiply the light years we found by the big number.
D'e,s'(Distance between earth and sun) = 1.496 * 10^13 centimeters<span>
D'e,*'(Distance between earth and Proxima) = </span><span>4.014 * 10^18 centimeters
</span>
Now we scale things down, we treat 1.496 * 10^13 centimeters as a SINGLE centimeter, because that's the distance between the earth and the sun. So all we have to do is divide (4.014 * 10^18 ) by (<span>1.496 * 10^13 ).
Why? because that how proportions work.

As a result, you get a mere 268335.7 centimeters.

To put that into perspective, that's only about 1.7 miles

A lot of my numbers came from google, so they are estimations and are not perfect, but its hard to be on really large scales.</span>
You might be interested in
Describe how the basic measurements of length, volume, and mass were determined in the
Anna11 [10]

the basic measurements of length, volume, and mass are 1 meter, 1 meter³  and 1kilogram.

One 10-millionth of the distance from the North Pole to the equator would equal one meter.

A cube with sides of one meter has a volume of one meter³.

The mass of 1,000 cubic centimeters of water is very close to (and was originally intended to be exactly) one kilograms.

but today a more precise way to define is available in which:

1 meter is defined as the distance traveled by light in 1/3×10⁸ sec.

The weight of a specific platinum-iridium prototype held by the International Bureau of Weights and Measures is defined as 1 kg.

moreover these measurements are man made and are not absolute and therefore many new units of measurement can be defined as long as they are fundamentally same everywhere.

learn more about metric system here:

brainly.com/question/2555957

#SPJ9

4 0
1 year ago
What is strain?
Sholpan [36]

Answer:

C) The ratio of the change in an object's length to its original length when stretched or compressed.

Explanation:

The formula for strain is:

Strain = Change in Length/Origin Length

Hence, it can be described as the percentile of change in the dimension with respect to the original dimension. So, whenever a tensile or a compressive force is applied on a body, its length changes. The ratio of this change to original length is called strain. So, the correct option is:

C) <u>The ratio of the change in an object's length to its original length when stretched or compressed.</u>

3 0
3 years ago
Consider a Hydrogen atom with the electron in the n 8 shell. What is the energy of this system? (The magnitude of the ground sta
Shtirlitz [24]

Answer:

The energy of an electron in the 8th shell is given by:  -0.2125 eV

The number of subshells is:  8

The number of orbitals is:  64

The number of electrons that fit on this shell is: 128

Explanation:

First, we find the energy of the electrons in the 8th shell. In order to do this, we recall that the energy of an electron (in the Hydrogen atom) whose principal number is n is given by:

E_{n}=-13.6\frac{1}{n^{2}}

Substituting n=8, we find that the energy is given by:

E_{8} = -13.6\frac{1}{8^{2}}=-0.2125

In order to find the number of subshells we recall that, for a given principal quantum number n, the possible values of the quantum number l, which corresponds to the number of subshells are:

0, 1, 2, ... , n-1

Since n = 8 in our problem, the possible values of l are: 0, 1, 2, 3, 4, 5, 6, 7. Therefore, the number of subshells are 8.

Now we continue with the number of orbitals. For every subshell l, we have 2l+1 possible values of m, which correspond to the orbitals. Since the possible values of l are: 0,1,2,3,4,5,6,7, therefore, we have to perform the sum:

\sum_{l=0}^{7}(2l+1) = 8^2=64

And we can conclude that the number of orbitals is equal to 64.

Finally, we know that we can fit two electrons per orbital, therefore we can have 64*2 = 128 electrons in the shell corresponding to n=8.

8 0
4 years ago
How much work do you do on a 15 N book in lifting it straight up for a distance of<br> 0.40 meters?
astra-53 [7]

Answer:

Work done, W = 6 J

Explanation:

It is given that,

Force of gravity acting on the book, weight of the book is 15 N

We need to find the work done in lifting the book straight up for a distance of  0.4 meters.

The weight of the book is acting in downward direction and the book is lifted straight up, it means angle between them is 180 degrees. Work done is given by :

W=Fd\cos180\\\\W=15\times 0.4\times \cos180\\\\W=-6\ J

So, the magnitude of work done in lifting the book is 6 joules.

7 0
3 years ago
Let surface S be the boundary of the solid object enclosed by x^2+z^2=4, x+y=6, x=0, y=0, and z=0. and, let f(x,y,z)=(3x)i+(x+y+
babunello [35]

a. I've attached a plot of the surface. Each face is parameterized by

• \mathbf s_1(x,y)=x\,\mathbf i+y\,\mathbf j with 0\le x\le2 and 0\le y\le6-x;

• \mathbf s_2(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf k with 0\le u\le2 and 0\le v\le\frac\pi2;

• \mathbf s_3(y,z)=y\,\mathbf j+z\,\mathbf k with 0\le y\le 6 and 0\le z\le2;

• \mathbf s_4(u,v)=u\cos v\,\mathbf i+(6-u\cos v)\,\mathbf j+u\sin v\,\mathbf k with 0\le u\le2 and 0\le v\le\frac\pi2; and

• \mathbf s_5(u,y)=2\cos u\,\mathbf i+y\,\mathbf j+2\sin u\,\mathbf k with 0\le u\le\frac\pi2 and 0\le y\le6-2\cos u.

b. Assuming you want outward flux, first compute the outward-facing normal vectors for each face.

\mathbf n_1=\dfrac{\partial\mathbf s_1}{\partial y}\times\dfrac{\partial\mathbf s_1}{\partial x}=-\mathbf k

\mathbf n_2=\dfrac{\partial\mathbf s_2}{\partial u}\times\dfrac{\partial\mathbf s_2}{\partial v}=-u\,\mathbf j

\mathbf n_3=\dfrac{\partial\mathbf s_3}{\partial z}\times\dfrac{\partial\mathbf s_3}{\partial y}=-\mathbf i

\mathbf n_4=\dfrac{\partial\mathbf s_4}{\partial v}\times\dfrac{\partial\mathbf s_4}{\partial u}=u\,\mathbf i+u\,\mathbf j

\mathbf n_5=\dfrac{\partial\mathbf s_5}{\partial y}\times\dfrac{\partial\mathbf s_5}{\partial u}=2\cos u\,\mathbf i+2\sin u\,\mathbf k

Then integrate the dot product of <em>f</em> with each normal vector over the corresponding face.

\displaystyle\iint_{S_1}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{6-x}f(x,y,0)\cdot\mathbf n_1\,\mathrm dy\,\mathrm dx

=\displaystyle\int_0^2\int_0^{6-x}0\,\mathrm dy\,\mathrm dx=0

\displaystyle\iint_{S_2}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{\frac\pi2}\mathbf f(u\cos v,0,u\sin v)\cdot\mathbf n_2\,\mathrm dv\,\mathrm du

\displaystyle=\int_0^2\int_0^{\frac\pi2}-u^2(2\sin v+\cos v)\,\mathrm dv\,\mathrm du=-8

\displaystyle\iint_{S_3}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^6\mathbf f(0,y,z)\cdot\mathbf n_3\,\mathrm dy\,\mathrm dz

=\displaystyle\int_0^2\int_0^60\,\mathrm dy\,\mathrm dz=0

\displaystyle\iint_{S_4}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{\frac\pi2}\mathbf f(u\cos v,6-u\cos v,u\sin v)\cdot\mathbf n_4\,\mathrm dv\,\mathrm du

=\displaystyle\int_0^2\int_0^{\frac\pi2}-u^2(2\sin v+\cos v)\,\mathrm dv\,\mathrm du=\frac{40}3+6\pi

\displaystyle\iint_{S_5}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^{\frac\pi2}\int_0^{6-2\cos u}\mathbf f(2\cos u,y,2\sin u)\cdot\mathbf n_5\,\mathrm dy\,\mathrm du

=\displaystyle\int_0^{\frac\pi2}\int_0^{6-2\cos u}12\,\mathrm dy\,\mathrm du=36\pi-24

c. You can get the total flux by summing all the fluxes found in part b; you end up with 42π - 56/3.

Alternatively, since <em>S</em> is closed, we can find the total flux by applying the divergence theorem.

\displaystyle\iint_S\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\iiint_R\mathrm{div}\mathbf f(x,y,z)\,\mathrm dV

where <em>R</em> is the interior of <em>S</em>. We have

\mathrm{div}\mathbf f(x,y,z)=\dfrac{\partial(3x)}{\partial x}+\dfrac{\partial(x+y+2z)}{\partial y}+\dfrac{\partial(3z)}{\partial z}=7

The integral is easily computed in cylindrical coordinates:

\begin{cases}x(r,t)=r\cos t\\y(r,t)=6-r\cos t\\z(r,t)=r\sin t\end{cases},0\le r\le 2,0\le t\le\dfrac\pi2

\displaystyle\int_0^2\int_0^{\frac\pi2}\int_0^{6-r\cos t}7r\,\mathrm dy\,\mathrm dt\,\mathrm dr=42\pi-\frac{56}3

as expected.

4 0
3 years ago
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