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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!:)

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Which type of current flows inside the flashlight A direct B alternating C repeating D pulsating

anzhelika [568]

The flashlight is powered by one or more batteries.
Batteries supply Direct current (DC) .

7 0

3 years ago

Read 2 more answers

An electron is projected with an initial speed of 5 3.2 10 / ⋅ m s directly toward a proton that is fixed in place. If the elect

Dvinal [7]

Answer:

The distance is $d =1.66*10^{-9}m$

Explanation:

From the question we are told that

The initial speed of the electron is $v_i = 3.2 *10^5 m/s$

The mass of electron is $m = 9.11*10^{-31}kg$

Let $d$ be the distance between the electron and the proton when the speed of the electron instantaneously equal to twice the initial value

Let $KE_i$ be the initial kinetic energy of the electron \

Let $KE_d$ be the kinetic energy of the electron at the distance $d$ from the proton

Considering that energy is conserved,

The energy at the initial position of the electron = The energy at the final position of the electron

i.e

$KE_i +U_1 = KE_d + U_2$

$U_1 \ and \ U_2$ are the potential energy at the initial position of the electron and at distance d of the electron to the proton

Here $U_1 = 0$

So the equation becomes

$\frac{1}{2} mv_i^2 = \frac{1}{2} mv_d^2 + \frac{kq_1 q_2}{d}$

Here $q_1 \ and \ q_2$ are the charge on the electron and the proton and their are the same since a charge on an electron is equal to charge on a proton

$k$ is electrostatic constant with value $8.99*10^9 N \cdot m^2 /C^2$

i.e $q = 1.602 *10^{-19}C$

$v_d$ is the velocity at distance d from the proton = 2 $v_i$

So the equation becomes

$\frac{1}{2}mv_i^2 = \frac{1}{2} m (2v_i)^2 -\frac{k(q)^2}{d}$

$\frac{1}{2} mv_i^2 = 4 [\frac{1}{2}mv_i^2 ]- \frac{k(q)^2}{d}$

$3[\frac{1}{2}mv_i^2 ] = \frac{k(q)^2}{d}$

Making d the subject of the formula

$d = \frac{2k(q)^2}{3mv_i^2}$

$= \frac{2* 8.99*10^9 *(1.602*10^{-19}^2)}{3 * 9.11*10^{-31} *(3.2*10^5)^2}$

$=1.66*10^{-9}m$

6 0

3 years ago

Aluminum and copper are good ____ because they don't have a strong hold on valence electrons. * a. Insulators b. Transistors c.

SOVA2 [1]

aluminum and copper are C.) conductors

6 0

3 years ago

a rocket with a mass of 4kg accelerates up from the ground at a rate of 20ft m/s^2, what is the drag force acting on the rocket

Maru [420]

The drag force acting on the rocket is 80N.

<h3>Give an explanation of drag force?</h3>

The divergence in velocity between the fluid and the item, also known as drag, exerts a force on it. Between the liquid and the solid object, there should be motion. Drag is absent in the absence of motion.

The air molecules are more compressed (pushed together) on the surfaces that are facing the front while being more dispersed (spread out) on the surfaces facing the back. Turbulent flow, which occurs when air layers split from the surface and start to swirl, is what causes this.

The drag force acting on the rocket F = ma

Given,

m = 4kg, a = 20ftm/s²

Substituting m and a values in the above formula,

The drag force acting on the rocket F = 4×20

The drag force acting on the rocket F = 80N.

To know more about drag force visit:

brainly.com/question/15144984

#SPJ4

8 0

1 year ago

As a consultant to the soft-drink industry, Dr. J is given the task of conducting the ultimate Pepsi taste test. This is Dr. J's

dem82 [27]

Answer:

A) How much work does the Pepsi do on the bullet = 0.0625J

B) At what velocity does the Pepsi hit the floor = 7.67m/s

Explanation:

Given mass of bullet = 5g

initial velocity = 500 m/sec

final velocity = 5 m/sec

From work done = Force X Distance

WD = Mad

a = v - u /t

Workdone ; m(v - u )/t

A) The work done by the Pepsi is equal to the change in kinetic energy while in the Pepsi:

but The work done by the container wall on the bullet is equal to the change in kinetic energy on either side of the wall: W = change in kinetic energy = 1/2 mv²

= 1/2 X 0.005 x 500² - 1/2 X 0.055 x 5²

= 624.94J

Therefore ; The work done by the Pepsi is equal to the change in kinetic energy while in the Pepsi = W = change in KE = 0.0625 - 0 = 0.0625J

B) At what velocity does the Pepsi hit the floor?

From conservation of energy principle; PE = KE

mgh = 1/2mv²

0.98 X 3 = 1/2v²

velocity = 7.67m/s

3 0

3 years ago