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3 years ago

Which of these is an example of interactions between the atmosphere and biosphere?

Physics

2 answers:

Alex17521 [72]3 years ago

8 0

Answer: d. uprooting of trees during tornadoes.

Explanation:

An atmosphere can be define as the mass of gases collectively called as air. It covers the earth and protects it from the direct incident rays coming from the sun which can cause damage to the living beings.

An biosphere is the important sphere of the earth which consists of living beings.

A tornado is a natural disaster which is a strong wind storm. It consists of strong column of air that spins very fast and causes a lot of damage to the material which comes in it's direct contact.

Uprooting of trees during tornado is the correct option because of the fact that tornado initially develop in the atmosphere and it effects the biosphere. The tree is the part of the biosphere.

SashulF [63]3 years ago

6 0

It might be D, Tornadoes deal with atmosphere affecting trees in the biosphere

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Answer:

A. Aristotle

Explanation:

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Water is being boiled in an open kettle that has a 0.52-cm-thick circular aluminum bottom with a radius of 12.0 cm. If the water

tangare [24]

Answer:

$T_b=107.3784\ ^{\circ}C$

Explanation:

Given:

thickness of the base of the kettle, $dx=0.52\ cm=5.2\times 10^{-3}\ m$

radius of the base of the kettle, $r=0.12\ m$

temperature of the top surface of the kettle base, $T_t=100^{\circ}C$

rate of heat transfer through the kettle to boil water, $\dot Q=0.409\ kg.min^{-1}$

We have the latent heat vaporization of water, $L=2260\times 10^3\ J.kg^{-1}$

and thermal conductivity of aluminium, $k=240\ W.m^{-1}.K^{-1}$

So, the heat rate:

$\dot Q=\frac{0.409\times 2260000}{60}$

$\dot Q=15405.67\ W$

From the Fourier's law of conduction we have:

$\dot Q=k.A.\frac{dT}{dx}$

$\dot Q=k\times \pi.r^2\times \frac{T_b-T_t}{5.2\times 10^{-3}}$

where:

$A=$ area of the surface through which conduction occurs

$T_b=$ temperature of the bottom surface

$15405.67=240\times \pi\times 0.12^2\times \frac{T_b-100}{5.2\times 10^{-3}}$

$T_b=107.3784\ ^{\circ}C$ is the temperature of the bottom of the base surface of the kettle.

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3 years ago

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