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Sonja [21]
2 years ago
11

If our Sun and solar system had formed at the same time as the very first stars in the universe, describe how and why our solar

system would be different than the one we have today and the impact this would likely have had on the formation of life. (You will need to consider the state of the universe at the time of the formation of the first stars.
Physics
1 answer:
aleksley [76]2 years ago
8 0

Answer:

How: It will have a very different chemical composition.

Why: Because the interstellar medium at the beginning of universe were compounded only of Hydrogen and Helium.

Explanation:

Once the Big Bang passed and the universe began its expansion, the temperature started to decrease, what allowed the combination of hydrogen and helium. Therefore, the chemical composition of the molecular cloud in which the first stars were formed only held those elements.

If the Sun and the solar system were formed from a molecular cloud with that composition, it will both have an absence of metallic elements (elements heavier than helium), since all of them are formed through nuclear reactions in the core of the stars, and each time a star dies, the interstellar medium for the next generation is enriched.

Key elements for life as Carbon, Nitrogen, Oxygen, Phosphorus and Sulfur will be missing for that hypothetical scenario.

For example, the lack of O3 (ozone) in the atmosphere of earth will allow that ultraviolet light hits the earth surface.

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An object is 6.0 cm in front of a converging lens with a focal length of 10 cm.Use ray tracing to determine the location of the
Masja [62]

As we know that

\frac{1}{d_i} + \frac{1}{d_0} = \frac{1}{f}

here we know that

d_0 = 6 cm

f = 10 cm

now from above equation we have

\frac{1}{d_i} + \frac{1}{6} = \frac{1}{10}

d_i = -15 cm

so image will form on left side of lens at a distance of 15 cm

This image will be magnified and virtual image

Ray diagram is attached below here

8 0
2 years ago
A roller coaster, traveling with an initial speed of 15 meters per second, decelerates uniformly at â7.0 meters per second2 to a
Harrizon [31]
<span>By algebra, d = [(v_f^2) - (v_i^2)]/2a. Thus, d = [(0^2)-(15^2)]/(2*-7) d = [0-(225)]/(-14) d = 225/14 d = 16.0714 m With 2 significant figures in the problem, the car travels 16 meters during deceleration.</span>
8 0
2 years ago
The maximum tension that a 0.80 m string can tolerate is 15 N. A 0.35-kg ball attached to this string is being whirled in a vert
zimovet [89]

Answer:

v=5.86 m/s

Explanation:

Given that,

Length of the string, l = 0.8 m

Maximum tension tolerated by the string, F = 15 N

Mass of the ball, m = 0.35 kg

We need to find the maximum speed the ball can have at the top of the circle. The ball is moving under the action of the centripetal force. The length of the string will be the radius of the circular path. The centripetal force is given by the relation as follows :

F=\dfrac{mv^2}{r}

v is the maximum speed

v=\sqrt{\dfrac{Fr}{m}} \\\\v=\sqrt{\dfrac{15\times 0.8}{0.35}} \\\\v=5.86\ m/s

Hence, the maximum speed of the ball is 5.86 m/s.

3 0
2 years ago
How would you solve for x I cant remember right now 4x+6x=9x-10
-Dominant- [34]
Combine all of the x's on one side of the equation and then finish the problem!
8 0
2 years ago
Read 2 more answers
A 3kW oven supplied with 9mJ of energy.How many minutes can it run for?
andreev551 [17]

<span>Hello!
 
We have the following data:
</span>
Time (T) = ? (in minutes)
Power (P) = 3 kW → 3000 W
Energy (E) = 9 MJ → 9000000 J or (W/s)

Formula of the consumption of electric energy:

P =  \frac{E}{T}

Solving:

P = \frac{E}{T}

P = \frac{E}{T} \to T =  \frac{E}{P}

T =  \frac{9000\diagup\!\!\!\!0\diagup\!\!\!\!0\diagup\!\!\!\!0\:\diagup\!\!\!\!W/s}{3\diagup\!\!\!\!0\diagup\!\!\!\!0\diagup\!\!\!\!0\:\diagup\!\!\!\!W}

\boxed{T = 3000\:seconds}

How many minutes can it run for? (<span>Let's convert in minutes)
</span>
1 minute --------- 60 seconds
y minute --------- 3000 seconds

\frac{1}{y} = \frac{60}{3000}

<span>Product of extremes equals product of means
</span>
60*y = 1*3000

60y = 3000

y =  \frac{3000}{60}

\boxed{\boxed{y = 50\:minutes}}\end{array}}\qquad\quad\checkmark


I hope this helps! =)
<span>

</span>
7 0
2 years ago
Read 2 more answers
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