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padilas [110]
3 years ago
12

Why are oxygen and hydrogen two of the most abundant elements in living things

Physics
1 answer:
katen-ka-za [31]3 years ago
6 0

Answer:

Oxygen and Hydrogen are the most abundant elements found in living beings because of their combination to form water together that builds the 70% of the living body, oxygen being used in respiration and Hydrogen being used to form other compounds as it has the simplest structure and least requirements to bond with other elements.

Explanation:

To build something, it's is important to start with the simplest formation and proceed to the complex ones. Same case applies on living beings. With the creation of all living bodies, two of the most abundant elements Hydrogen and Oxygen tends to volunteer in the formation of living bodies. Their simplest structure and least requirements to bond easily with other elements also make them suitable for the creation of living bodies.

Hydrogen consisting of only one proton and one neutron, easily collaborates with other elements and forms the complex compounds. Whereas, oxygen being processed under the respiration system, helps in developing important organs and compounds. And, most importantly not to forget point is that they both combine to make the significant compound of the living body i.e water. However, Carbon, Nitrogen and other traces are also present in the living beings but these two serve as the saviors of the war.

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Which event indicates an endothermic chemical reaction? HELP ME ASAP PLZ!!!!
Mars2501 [29]

Answer: C

Explanation:

3 0
3 years ago
An engine absorbs 1.69 kJ from a hot reservoir at 277°C and expels 1.25 kJ to a cold reservoir at 27°C in each cycle.
Anna35 [415]

Answer

Given,

Energy absorbed, Q_H = 1.69\ kJ

Energy expels,Q_C =  1.25\ kJ

Temperature of cold reservoir, T = 27°C

a) Efficiency of engine

 \eta = \dfrac{Q_H - Q_C}{Q_H}\times 100

 \eta = \dfrac{1.69 - 1.25}{1.69}\times 100

\eta =26.03 %

b) Work done by the engine

 W = Q_H- Q_C

 W =1.69 - 1.25

 W = 0.44\ kJ

c) Power output

     t = 0.296 s

   P = \dfrac{W}{t}

   P = \dfrac{0.44}{0.296}

   P = 1.486\ kW

8 0
3 years ago
A mass M is attached to an ideal massless spring. When this system is set in motion with amplitude A, it has a period T. What is
Tasya [4]

Answer:

C) T

Explanation:

M = Mass attached to an ideal spring

A = Amplitude of the motion

T = Time period of oscillation

k = Spring constant of the spring

A = Amplitude of the motion

Time period of oscillation of the mass attached to the spring is given as

T = 2\pi \sqrt{\frac{M}{k} }

So we see that the time period does not depend on the amplitude. hence the period of oscillation remains the same.

8 0
3 years ago
A tank, shaped like a cone has height 12 meter and base radius 1 meter. It is placed so that the circular part is upward. It is
Temka [501]

Answer:

376966.991 Joules

Explanation:

Given that :

the height = 12  m

Let assume the tank have a thickness  = dh

The radius of the tank by using the concept of similar triangle is :

\dfrac{1}{r} = \dfrac{12}{h}

r = \dfrac{h}{12}

The area of the tank = \mathbf{\pi r^2}

The area of the tank = \mathbf{\pi( \dfrac{h}{12})^2}

The area of the tank = \mathbf{ \dfrac{\pi}{144}h^2}

The volume of the tank is  = area × thickness

= \mathbf{ \dfrac{\pi}{144}h^2 \  dh}

Weight of the element = \rho_ g * volume

where;

\rho_g = density of water ; which is given as 10000 N/m³

So;

Weight of the element = \mathbf{ 10000 *\dfrac{\pi}{144}h^2 \  dh}

Weight of the element = \mathbf{69.44 \ \pi  \ h^2 \  dh}

However; the work required to pump this water = weight × height  rise

where the height rise = 12 - h

the work required to pump this water  = \mathbf{69.44 \ \pi  \ h^2 \  dh}(12 - h)

the work required to pump this water  = \mathbf{69.44 \pi (12h^2-h^3)dh}

We can determine the total workdone by integrating the work required to pump this water

SO;

Workdone = \mathbf{\int\limits^{12}_0 {69.44 \pi(12h^2-h^3)} dh}

= \mathbf{ 69.44 \pi \int\limits^{12}_0 {(12h^2-h^3)} dh}

=  \mathbf{ 69.44 \pi[ \frac{12h^3}{3}-  \frac{h^4}{4}]^{12}}_0} }

= \mathbf{69.44 \pi [ \frac{12^4}{3}-\frac{12^4}{4}]}

= \mathbf{69.44 \pi*12^4 [ \frac{4-3}{12}]}

= \mathbf{69.44 \pi*12^4 *\frac{1}{12}}

= 376966.991 Joules

6 0
3 years ago
Pls answer!! 20 points and will give crown to best answer!!
docker41 [41]
I am 99999.999999999% sure that it is his second law of motion. 

Hope this helped! :) Please mark brainliest 
5 0
4 years ago
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