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3 years ago

What changes were necessary in earth’s atmosphere to allow complex organism to evolve

Physics

2 answers:

jok3333 [9.3K]3 years ago

6 0

Answer:

Earth’s original atmosphere was rich in methane, ammonia, water vapor, and the noble gas neon, but it lacked free oxygen.Oxygen had to become a major atmospheric gas and the ozone layer had to develop.

Zepler [3.9K]3 years ago

5 0

A collision from above such as a meteor and an increase in volcano activity.

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A block of mass 0.221 kg is placed on top of a light, vertical spring of force constant 5365 N/m and pushed downward so that the

Anvisha [2.4K]

Answer:

The maximum height above the point of release is 11.653 m.

Explanation:

Given that,

Mass of block = 0.221 kg

Spring constant k = 5365 N/m

Distance x = 0.097 m

We need to calculate the height

Using stored energy in spring

$U=\dfrac{1}{2}kx^2$ ...(I)

Using gravitational potential energy

$U' =mgh$ ....(II)

Using energy of conservation

$E_{i}=E_{f}$

$U_{i}+U'_{i}=U_{f}+U'_{f}$

$\dfrac{1}{2}kx^2+0=0+mgh$

$h=\dfrac{kx^2}{2mg}$

Where, k = spring constant

m = mass of the block

x = distance

g = acceleration due to gravity

Put the value in the equation

$h=\dfrac{5365\times(0.097)^2}{2\times0.221\times9.8}$

$h=11.653\ m$

Hence, The maximum height above the point of release is 11.653 m.

3 0

3 years ago

escribe the differences between the terrestrial and jovian planets and classify each of the 8 planets to the proper type

Radda [10]

Answer:

*******

Explanation:

is obviously the correct answer they are both so different yet so alike

6 0

2 years ago

Read 2 more answers

Which form of

Soloha48 [4]

I think it’s C b/c it works for me

3 0

2 years ago

An apple with a mass of 0.95 kilograms hangs from a tree branch 3.0 meters above the ground. If it falls to the ground, what is

lubasha [3.4K]

Ke = pe
pe = mgh
= 0.95 x 9.8 x 3
= 27.93 J

6 0

3 years ago

Read 2 more answers

a system of four particles moves along one dimension. the center of mass of the system is at rest, and the particles do not inte

Zinaida [17]

The velocity of the particle 4 at time, t = 2.83 s, is -14.1 m/s.

The given parameters;

m1 = 1.45 kg, v1(t) = (6.09m/s) + (0.299m/s^2) × t

m2 = 2.81 kg, v2(t) = (7.83m/s) + (0.357m/s^2) × t

m3 = 3.89 kg, v3(t) = (8.09m/s) + (0.405m/s^2) × t

m4 = 5.03kg

The velocity of the center mass of the particles is calculated as;

McmVcm = m1v1 + m2v2 +m3v3+m4v4

Vcm= m1v1 + m2v2 +m3v3 +m4v4/ Mcm

0 = m1v1 + m2v2 +m3v3 +m4v4/ Mcm

m1v1 + m2v2 +m3v3+m4v4 = 0

m4v4 = -(m1v1 + m2v2 +m3v3)

v4 =-(m1v1 + m2v2 +m3v3)/ m4

The velocity of particle 1 at time, t = 2.83 s;

vi = 6.09 + 0.299× 2.83

v1 = 6.94 m/s

The velocity of particle 3 at time, t = 2.83 s;

v2 = 7.83 + 0.357 × 2.83

v2 = 8.84 m/s

The velocity of particle 3 at time, t = 2.83 s;

v3 = 8.09 + 0.405 × 2.83

v3 = 9.24 m/s

The velocity of particle 3 at time, t = 2.83 s;

v4 = - (m1v1 + m2v2 + m3v3)/m4

v4 = -(1.45×6.94 + 2.81×8.84 + 3.89×9.24)/5.03

v4 = -14.4 m/s

Thus, the velocity of the particle 4 at time, t = 2.83 s, is -14.1 m/s.

Learn more about Velocity here:

brainly.com/question/18084516

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11 months ago