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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>
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Answer:

9.6m/s

Explanation:

Using the equation S=d/t where s=speed, d=distance, and t=time

plug in the known variables

S=120m/12.5s

S=9.6m/s

4 0
3 years ago
Match the events related to the formation of the universe with the stages during which they occurred.
igomit [66]

Answer:

does this even make sense

Explanation:

8 0
3 years ago
Mary starts at the edge of a circular platform that is slowly rotating on a frictionless axle. She then walks towards the opposi
bulgar [2K]

Answer:

Explanation:

The motion of Mary along the circular path is a centripetal.

As Mary moves from one edge of the circular platform to the other edge, she is covering a distance which is the radius of the circular path at a velocity.

According to the relationship

w = v/r where

w is the angular velocity

r is the radius

v is the linear velocity

Initially, before Mary starts, her linear speed is zero and her angular velocity is also zero. As she move towards the opposite edge, she is covering a distance of radius r. According to the formula, increase in radius will leads to decrease in her angular velocity and vice versa. As Mary starts moving towards the centre of the circular path, her angular velocity increases, at the centre of the platform, her angular velocity is at maximum at this point. As she moves further from the center to the other edge, her angular velocity decreases due to increase in distance covered across the circular path.

6 0
3 years ago
In our solar system, the most likely planet (other than Earth) to have life on it is currently thought to be
VLD [36.1K]

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6 0
3 years ago
Suppose that Hubble's constant were H0 = 51 km/s/Mly (which is not its actual value). What would the approximate age of the univ
bija089 [108]

Given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Given the data in the question;

Hubble's constant; H_0 = 51km/s/Mly

Age of the universe; t = \ ?

We know that, the reciprocal of the Hubble's constant ( H_0 ) gives an estimate of the age of the universe ( t ). It is expressed as:

Age\ of\ Universe; t = \frac{1}{H_0}

Now,

Hubble's constant; H_0 = 51km/s/Mly

We know that;

1\ light\ years = 9.46*10^{15}m

so

1\ Million\ light\ years = [9.46 * 10^{15}m] * 10^6 = 9.46 * 10^{21}m

Therefore;

H_0 = 51\frac{km}{\frac{s}{Mly} } = 51000\frac{m}{s\ *\ Mly}  \\\\H_0 = 51000\frac{m}{s\ *\ (9.46*10^{21}m)} \\\\H_0 =  5.39 *10^{-18}s^{-1}\\

Now, we input this Hubble's constant value into our equation;

Age\ of\ Universe; t = \frac{1}{H_0}\\\\t = \frac{1}{ 5.39 *10^{-18}s^{-1}} \\\\t = 1.855 * 10^{17}s\\\\We\ convert\ to\ years\\\\t =  \frac{ 1.855 * 10^{17}}{60*60*24*365}yrs \\\\t = \frac{ 1.855 * 10^{17}}{31536000}yrs\\\\t = 5.88 *10^9 years

Therefore, given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Learn more: brainly.com/question/14019680

6 0
3 years ago
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