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Citrus2011 [14]
3 years ago
6

The oxygen in earth’s atmosphere is a product of what?

Physics
1 answer:
Nina [5.8K]3 years ago
7 0

Answer:

The answer is tiny organisms known as cyanobacteria, or blue-green algae. These microbes conduct photosynthesis: using sunshine, water and carbon dioxide to produce carbohydrates and, yes, oxygen.

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Explain Thomsons model of an atom<br><br><br>please its aurgent fast​
Aleks04 [339]

Answer:

Thomson's model showed an atom that had a positively charged medium, or space, with negatively charged electrons inside the medium. After its proposal, the model was called a "plum pudding" model because the positive medium was like a pudding, with electrons, or plums, inside.

3 0
2 years ago
How much time will it take to perform 440 joule of work at a rare of 11 w?​
kifflom [539]

Answer:

40sec

Explanation:

Data

Work = 440 J

Power= 11watt

time = ?

Power = work done/time

===> time = work done/power

= 440/11

= 40sec

7 0
2 years ago
A satellite completes one revolution of a planet in almost exactly one hour. At the end of one hour, the satellite has traveled
topjm [15]

average velocity is vector displacement / time

time is "almost exactly one hour"

disp = -10m

v= -10/1x60x60 = -1/360m/s

5 0
3 years ago
How energy and power are different ???
Naya [18.7K]

Answer:

Explanation:

Energy is what makes change happen and can be transferred form one object to another. ... Power is the rate at which energy is transferred. It is not energy but is often confused with energy. The watt is the most commonly used unit of measure for power.

6 0
2 years ago
A spherical asteroid of average density would have a mass of 8.7×1013kg if its radius were 2.0 km. 1. If you and your spacesuit
Law Incorporation [45]

1. 0.16 N

The weight of a man on the surface of asteroid is equal to the gravitational force exerted on the man:

F=G\frac{Mm}{r^2}

where

G is the gravitational constant

M=8.7\cdot 10^{13}kg is the mass of the asteroid

m = 100 kg is the mass of the man

r = 2.0 km = 2000 m is the distance of the man from the centre of the asteroid

Substituting, we find

F=(6.67\cdot 10^{-11}m^3 kg^{-1} s^{-2})\frac{(8.7\cdot 10^{13} kg)(110 kg)}{(2000 m)^2}=0.16 N

2. 1.7 m/s

In order to stay in orbit just above the surface of the asteroid (so, at a distance r=2000 m from its centre), the gravitational force must be equal to the centripetal force

m\frac{v^2}{r}=G\frac{Mm}{r^2}

where v is the minimum speed required to stay in orbit.

Re-arranging the equation and solving for v, we find:

v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{(6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2})(8.7\cdot 10^{13} kg)}{2000 m}}=1.7 m/s

3 0
3 years ago
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