All categories

3 years ago

Select all of the answers that apply.

2 answers:

6 0

The answer is letter A. meteorite bombardment.
During the Earth's earliest beginning, it went through a period of catastrophic and intense formation. By 3-8- 4.1 billion years ago, Earth's atmosphere was never the same as today. This is because of its formation during the pre-Cambrian period whereby t<span>he Earth formed under so much heat and pressure that it formed as a molten planet.
</span>

Earth was bombarded continuously by the remnants of the dust and debris — like asteroids, meteors and comets — until it formed into a solid sphere, pulled into orbit around the sun and began to cool down during the Hellish period (4.5 to 3.8 billion years ago).

<span> </span>.To know more of this topic, see attached file.

Download docx

Helga [31]3 years ago

4 0

Answer: meteorite bombardment

Explanation: This is because of its formation during the pre-Cambrian period whereby the Earth formed under so much heat and pressure that it formed as a molten planet.

You might be interested in

An automobile and train move together along parallel paths at 23.2 m/s. The automobile then undergoes a uniform acceleration of

lara [203]

Answer:

Train will 588.215 m ahead of automobile

Explanation:

We have given initially speed of both train and automobile is same as 23 m/sec

After that acceleration of automobile $a=-4m/sec^2$

Initial velocity u = 23 m/sec

Final velocity v = 0 ( as it comes to rest )

According to first law of motion

v=u+at

$0=23-4\times t$

t = 5.75 sec

Now according to third law of motion $v^2=u^2+2as$

$0^2=23^2-2\times 4\times s$

s = 66.125 m

Now automobile accelerates at with acceleration $a=2.09m/sec^2$

Initial velocity u = 23 m/sec

Final velocity v = 23.2 m/sec

So $time\ t =\frac{v-u}{a}=\frac{23.2-23}{2.09}=0.0956sec$

Distance traveled by automobile in 0.0956 sec

$s=ut+\frac{1}{2}at^2=23\times 0.0956+\frac{1}{2}\times 2.09\times 0.0956^2=2.208m$

So total distance traveled by automobile = 66.125+2.208 = 68.333 m

Total time = 5.75 +22.7+0.0956=28.545 sec

Sp distance traveled by train in 28.545 sec

= 28.545×23 = 656.548 m

So distance between automobile and train = 656.548 - 68.33=588.215 m

So train will 588.215 m ahead of automobile

3 0

3 years ago

A block is projected up a frictionless inclined plane with initial speed v0 = 7.14 m/s. The angle of incline is θ = 36.5°. (a) H

Wewaii [24]

Answer:

$(a)x=4.37m\\\\(b)t=1.225s\\\\(c) v_{f}=7.14m/s$

Explanation:

Given data

$v_{o}=7.14m/s\\\alpha =36.5^o$

For Part (a)

Starting with the -ve acceleration of the body (opposite to the gravitational force)

$a=-gSin\alpha \\a=-(9.8m/s^2)Sin(36.5)\\a=-5.83m/s^2$

Using equation of motion

$v_{f}^2=v_{o}^2+2ax\\(0m/s)^2=(7.14m/s)^2+2(-5.83m/s^2)x\\-(7.14m/s)^2=2(-5.83m/s^2)x\\x=\frac{-(7.14m/s)^2}{2(-5.83m/s^2}\\ x=4.37m$

For Part (b)

Using the result in Part (a) we can substitute in other equation of motion to get time t:

$x=\frac{1}{2}vt\\ 4.37m=\frac{1}{2}(7.14m/s)t\\ (7.14m/s)t=2*(4.37)\\t=8.744/7.14\\t=1.225s$

For Part (c)

At state 2 where vo=0m/s and the acceleration is positive (same direction as the gravitational force)

$a=gSin\alpha \\a=(9.8m/s^2)Sin(36.5)\\a=5.83m/s^2\\\\\\v_{f}^2=v_{o}^2+2ax\\v_{f}^2=(0m/s)^2+2(5.83m/s^2)(4.37m)\\v_{f}^2=50.95\\v_{f}=\sqrt{50.95}\\ v_{f}=7.14m/s$

4 0

3 years ago

Which of these is transparent?

jolli1 [7]

The answer is definitely A. A transparent material is a material that allows light to pass through it and can be seen through. A glass jar can be seen through. Some people might say B, which is a mirror, but a mirror cannot be seen through, it only shows reflection. Hope i helped. Have a nice day and Happy New Year.

3 0

3 years ago

Read 2 more answers

Standing waves of frequency 68 hz are produced on a string that has mass per unit length 0.0160 kg/m. with what tension must the

mario62 [17]

0.54 ghost I think probably

3 0

3 years ago

A car accelerates uniformly from rest to a speed of 23.7m/s in 6.5 seconds. Find the distance the

frez [133]

The car's average speed during that time is (23.7/2) = 11.85 m/s .

Distance = (11.85 m/s) x (6.5 sec) = 77.025 meters .

5 0

4 years ago