All categories

3 years ago

Does fire have ways to protect itself from changes in the environment

Physics

1 answer:

Alexandra [31]3 years ago

6 0

Answer:

I think no because the critical properties distinguishing life is adaptation to changing environment and self replication of the information encoding the life process. Fire does not change its process to adapt to its environment, e.g. moving toward more fuel or storing and conserving fuel when it is in short supply.

Hope this help you!:)

You might be interested in

Which statement accurately describes what happens when water vapor condenses into dew in terms of energy?

son4ous [18]

The correct explanation for the process of formation of dew from the water vapour is The water releases energy which causes the water molecules to have less kinetic and potential energy, changing their configuration from gas to liquid.

vapour is a gaseous form, while dew is a liquid state of matter therefore it's clear that vapour has to release energy to come in the liquid state, and in this process vapour, water molecules lose their kinetic and potential energy.

6 0

3 years ago

Read 2 more answers

What does a mass extinction look like in the fossil record?

IrinaK [193]

Answer:

the answer is D

Explanation:

Mass extinctions were first identified by the obvious traces they left in the fossil record. ... Such dramatic changes in adjacent rock layers make it clear that mass extinctions were geologically rapid and suggest that they were caused by catastrophic events (e.g., a period of intense volcanic activity).

8 0

3 years ago

Read 2 more answers

The electric potential inside a charged spherical conductor of radius R is is given by V = keQ/R, and the potential outside is g

Bumek [7]

Answer:

For outer points of shell

$E = \frac{k_eQ}{r^2}$

Now for inner point of shell

$E = 0$

Explanation:

As we know that out side the shell electric potential is given as

$V = \frac{K_e Q}{r}$

inside the shell the electric potential is given as

$V = \frac{K_e Q}{R}$

now we know the relation between electric potential and electric field as

$E = - \frac{dV}{dr}$

so we can say for outer points of the shell

$E = -\frac{dV}{dr}$

$E = - \frac{d}{dr}(\frac{K_eQ}{r})$

$E = \frac{k_eQ}{r^2}$

Now for inner point again we can use the same

$E = - \frac{dV}{dr}$

$E = - \frac{d}{dr}(\frac{K_eQ}{R})$

$E = 0$

6 0

3 years ago

Balanced or Unbalanced?

Galina-37 [17]

Unbalanced because if it is pushing then stopping, that means that it is unbalanced.

6 0

3 years ago

Razona si, en las experiencias de Faraday, el efecto producido moviendo el circuito será el mismo que el producido moviendo el i

const2013 [10]

Answer:

moving the circuit or the magnet gives the same result

Explanation:

The faraday effect establishes that the temporal variation of imaginative flow produces an electric potential

fem = $\frac{d \phi }{dt}$ dfi / dt

the magnetic flux is

Ф = B. A = B A cos θ

suppose for simplicity that the angle is zero so cos 0 = 1

Φ = B A

By analyzing this expression, the change in magnetic flux can converge while keeping the magnetic field fixed and varying the electric field or keeping the electric field fixed and varying the magnetic field.

Consequently moving the circuit or the magnet gives the same result

5 0

3 years ago