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3 years ago

How did photosynthesis start?

Physics

2 answers:

aleksklad [387]3 years ago

6 0

Answer:

Fossils of what are thought to be filamentous photosynthetic organisms have been dated at 3.4 billion years old. More recent studies, reported in March 2018, also suggest that photosynthesis may have begun about 3.4 billion years ago.

ycow [4]3 years ago

3 0

Answer:

Overwhelming evidence indicates that eukaryotic photosynthesis originated from endosymbiosis of cyanobacterial-like organisms, which ultimately became chloroplasts (Margulis, 1992). So the evolutionary origin of photosynthesis is to be found in the bacterial domain.

Explanation:

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A hydrogen atom in a galaxy moving with a speed of 6.65×106 m/???? away from the Earth emits light with a wavelength of 5.13×10−

Mumz [18]

Answer:

The observed wavelength on Earth from that hydrogen atom is $5.24\times 10^{-7}\ m$ .

Explanation:

Given that,

The actual wavelength of the hydrogen atom, $\lambda_a=5.13\times 10^{-7}\ m$

A hydrogen atom in a galaxy moving with a speed of, $v=6.65\times 10^6\ m/s$

We need to find the observed wavelength on Earth from that hydrogen atom. The speed of galaxy is given by :

$v=c\times \dfrac{\lambda_o-\lambda_a}{\lambda_a}$

$\lambda_o$ is the observed wavelength

$\lambda_o=\dfrac{v\lambda_a}{c}+\lambda_a\\\\\lambda_o=\dfrac{6.65\times 10^6\times 5.13\times 10^{-7}}{3\times 10^8}+5.13\times 10^{-7}\\\\\lambda_o=5.24\times 10^{-7}\ m$

So, the observed wavelength on Earth from that hydrogen atom is $5.24\times 10^{-7}\ m$ . Hence, this is the required solution.

8 0

3 years ago

Johnson made a hole at the bottom of a plastic bottle containing water. However, he noticed that the water did not flow out from

Sphinxa [80]

Answer:

a) See explanation below

b) At X.

Explanation:

Please, see the picture attached with the image of the plastic bottle for this question.

<u>(a) Explain why the water could not flow out of the bottle.</u>

What makes the water flow out of the botlle is the force of gravity, whic attracts the water towards the Earth.

When Johnson made a small hole at the bottom of the plastic bottle containing water, the air outside the bottle, which surrounds it and exerts a pressure all over the outer walls of the bottle, exerted a force against the small area of water "over" the hole that is in contact with the air.

Thus, this force of the air pushing upward through the wall opposed the force of gravity pulling downward making the net force zero and the water cannot fall.

<u>(b) To make the water flow out more easily, his teacher suggested making another hole. At which position - X, Y or Z, should he make the 2nd hole in order for the water to flow out the fastest?</u>

You must open the hole at a place where there is not water but air, such that the outer air can enter in the bottle.

That will make that the pressure in the space over the water inside the bottle be equal to the pressure outside.

The pressure of the air above the water will push it downward. Now, the force from the pressure of air inside the water, which is downward, opposes the upward force from the pressure of air around the first hole, and the net force will be downward, making the water flow out more easily.

Thus, the position where he should make the second hole in order for the water to flow fastest is at X.

4 0

3 years ago

A piano wire with mass 2.95 g and length 79.0 cm is stretched with a tension of 29.0 N . A wave with frequency 105 Hz and amplit

Romashka-Z-Leto [24]

The concept needed to solve this problem is average power dissipated by a wave on a string. This expression ca be defined as

$P = \frac{1}{2} \mu \omega^2 A^2 v$

Here,

$\mu$ = Linear mass density of the string

$\omega =$ Angular frequency of the wave on the string

A = Amplitude of the wave

v = Speed of the wave

At the same time each of this terms have its own definition, i.e,

$v = \sqrt{\frac{T}{\mu}} \rightarrow$ Here T is the Period

For the linear mass density we have that

$\mu = \frac{m}{l}$

And the angular frequency can be written as

$\omega = 2\pi f$

Replacing this terms and the first equation we have that

$P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2(\sqrt{\frac{T}{\mu}})$

$P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2 (\sqrt{\frac{T}{m/l}})$

$P = 2\pi^2 f^2A^2(\sqrt{T(m/l)})$

PART A ) Replacing our values here we have that

$P = 2\pi^2 (105)^2(1.8*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})$

$P = 0.2320W$

PART B) The new amplitude A' that is half ot the wavelength of the wave is

$A' = \frac{1.8*10^{-3}}{2}$

$A' = 0.9*10^{-3}$

Replacing at the equation of power we have that

$P = 2\pi^2 (105)^2(0.9*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})$

$P = 0.058W$

8 0

3 years ago

Three examples that prove newton's law of motion

s344n2d4d5 [400]

If you have a skateboard and you skate into a tree on accident the same amount of force you put onto that tree when you was on the skateboard will come back at you when you bounce back

7 0

3 years ago

Which of the following is the best definition of vaporization

Vinvika [58]

Change of a liquid to a gas

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3 years ago