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3 years ago

How were the fossil symbols and mountain bets helpful in deciding where to move the continenets

1 answer:

6 0

<span>Similar fossils found on different continents helped geologists determine how the continents used to be connected. Mountain belts marked the boundaries of moving plates, which showed in which direction the different continents drifted. Extrapolating from this information, scientists had a rough idea of how the continents were arranged eons ago.</span>

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The force of gravity on a 1 kg object on the Earth's surface is approximately 9.8 N. For the same object in low-earth orbit arou

mojhsa [17]

Answer:

g = 8.61 m/s²

Explanation:

distance of the International Space Station form earth is 200 Km

mass of the object = 1 Kg

acceleration due to gravity on earth = 9.8 m/s²

mass of earth = 5.972 x 10²⁴ Kg

acceleration due to gravity = ?

r = 6400 + 200 = 6800 Km = 6.8 x 10⁶ n

using formula

$g = \dfrac{GM}{r^2}$

$g = \dfrac{6.67\times 10^{-11}\times 5.972\times 10^24}{(6.8\times 10^6)^2}$

g = 8.61 m/s²

3 0

3 years ago

a water-balloon launcher with mass 4 kg fires a 0.5 kg balloon with a velocity of 3 m/s to the east. what is the recoil velocity

kotykmax [81]

I think we will use the law of conservation of linear momentum;
M1V1 = M2V2
M1 = 4 kg (mass of the water balloon launcher)
V1=?
M2= 0.5 kg ( mass of the balloon)
V2 = 3 m/s

Therefore; 4 V1 = 0.5 × 3
4V1= 1.5
V1= 1.5/4
= 0.375 m/s

5 0

3 years ago

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Which of the following is an example of rotation

SOVA2 [1]

I believe it is D. Earth spinning on it's axis.

8 0

3 years ago

I tried to experiment of dropping candy into a liter of soda and caused a big explosion of foamy soda

Anestetic [448]

Answer:

.....what type of question is this

Explanation:

6 0

3 years ago

Two objects, Object A and Object B, need to be identified. Object A's index of refraction is determined to be 1.77, and Object B

Slav-nsk [51]

The correct answer is

C. Light can pass through Object B faster than it can pass through Object A.

In fact, the index of refraction of a material is defined as:

$n=\frac{c}{v}$

where c is the speed of light in vacuum and v is the speed of light in the material. Rearranging the equation, we can write the speed of light in the material as:

$v=\frac{c}{n}$

So we that, the smaller the refractive index n, the greater the speed of light in the material, v. In this problem, object B has lower refractive index than object A, so light travels faster in object B.

4 0

4 years ago

Read 2 more answers