1. Our solar system is the only place in the universe where gravity played a key part in the formation of planets.
2. Rocky planets are small, dense, and orbit relatively close to the sun, compared to the Jovian planets, which are large, less dense, and orbiting far from the sun.
3. _______
(1) The image of an object placed further from the lens than the focal point will be upside down and smaller than the object.
(2) When light rays reflect, they bounce back.
(3) Images formed by a concave lens will look magnified.
(4) When light rays enter a different medium, they bend.
<h3>
1.0 Object placed further from the lens than the focal point</h3>
The image of an object placed further from the lens than the focal point will be diminished and inverted.
Thus, the correct answer will be "upside down and smaller than the object".
<h3>2.0 What is reflection of light?</h3>
The ability of light to bounce back when it strike a hard surface is known as refection.
<h3>3.0 Image formed by concave lens</h3>
A concave lens is diverging lens is usually virtual, erect and magnified.
<h3>4.0 Refraction of light</h3>
The change in speed of light when it travels from medium to another medium is known as refraction. Refraction is also, the ability of light to bend around obstacles.
Learn more about reflection and refraction of light here: brainly.com/question/1191238
Answer:
Hydrogen and helium compounds.
Explanation:
We know that the solar System was formed around <u>4.6 billion years ago, </u>due to the gravitational collapse of a giant interstellar molecular cloud.
This cloud is a type of interstellar cloud and its density and size permit the formation of molecules, most commonly molecular hydrogen.
Therefore the principal substances were found before planets began to form are hydrogen and helium compounds, besides Rocks, metals, most of them in gaseous form.
I hope it helps you!
Answer:
the correct solution is 13 s
Explanation:
This is a kinematic problem, let's use accelerated rectilinear motion relationships.
For the first car it has an accelerometer of 2.0 m/s²
x = v₀₁ t + ½ a₁ t²
The second car leaves the same point, but 4.0 seconds later
x = v₀₂ (t-4) + ½ a₂ (t-4)²
With this form we use the same time for both cars.
The initial speeds are zero for both vehicles leave the rest, at the point where they are located has the same position
x = ½ a₁ t²
x = ½ a₂ (t-4)²
Let's solve
a₁ t² = a₂ (t-4)²
a₁/a₂ t² = t² -2 4 t + 16
t² (1- 2.0 / 4.0) - 8 t +16
t² 0.5 - 8 t +16 = 0
t² -16 t + 32 = 0
Let's solve the second degree equation
t = [16 ±√( 16² - 4 32)] / 2
t = ½ (16 ± 11,3)
Solutions
t1 = 13.66 s
t2 = 2.34 s
These are the mathematical solutions for the meeting point, but car 2 leaves after 4 seconds, so the only solution is 13.66 s
the correct solution is 13 s, if you have to select one the nearest 12s
