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IgorLugansk [536]
2 years ago
5

In 450 BCE, Greek philosopher Democritus first thought of the existence of tiny particles that compose everything around us. He

named them 'atomos', meaning _________.
A. None of these
B. Indivisible
C. Invisible
D. Particle
Physics
1 answer:
Masja [62]2 years ago
5 0

B. The meaning of 'atomos' according to Democritus in 450 BCE is indivisible.

<h3>The building block of every matter</h3>

A Greek philosopher  known as Democritus first thought of the existence of tiny particles that compose everything around us.

Democritus of Abdera, named the building blocks of matter atomos, meaning literally “indivisible.

Thus, the meaning of 'atomos' according to Democritus in 450 BCE is indivisible.

Learn more about atoms here: brainly.com/question/6258301

#SPJ1

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Which choice best characterizes K+ leakage channels? View Available Hint(s)
hoa [83]

Answer:

C. <u>Trans-membrane protein channels that are always open to allow K+ to cross the membrane without the additional input of energy</u>

Explanation:

As we know that K+ leakage channels indicate the Potassium leakage channel. The best characterizes about K+ leakage channels is that there is a Trans-membrane protein channel which is always open to allow K+ to cross the membrane without the additional input of energy.

Actually when a cell dies its membrane potential gets more positive and finally reaches zero .

The key point is that the cells spend energy to maintain the intra-cellular ionic concentrations constant.

The membrane permeability of K+ is much  higher than the membrane permeability of Na+.

Therefore, the correct choice is option (C).

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4 years ago
The moon has a mass of 7.34 . 102 kg and a radius of 1.74 . 106 meters. If you have a mass of 66 kg,
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Answer:

F=1.06\times 10^{-18}\ N

Explanation:

Given that,

Mass of the Moon, M=7.34\times 10^2\ kg

Mass of the person, m = 66 kg

The radius of Moon, r=1.74\times 10^6\ m

We need to find the force between the person and the Moon. The formula for the gravitational force is given by :

F=G\dfrac{Mm}{r^2}\\\\F=6.67\times 10^{-11}\times \dfrac{7.34\times 10^2\times 66}{(1.74\times 10^6)^2}\\\\=1.06\times 10^{-18}\ N

So, the required force is 1.06\times 10^{-18}\ N.

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