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Tanzania [10]
3 years ago
8

Why can’t we just stop producing Co2

Physics
1 answer:
koban [17]3 years ago
3 0

What a great idea !  Sadly, we don't know how.

CO₂ is produced by the process of burning almost everything we burn for energy ... like coal, natural gas, oil, paper, wood, gasoline, alcohol, candle-wax, airplane fuel, whale oil, olive oil, Diesel fuel . . . stuff like that.

We certainly could stop producing CO₂ if we're willing to stop driving most cars, buses, taxis, trains, and airplanes, stop generating most of the electricity we use, stop heating most homes and other buildings, stop doing most of the cooking that we do, stop making most of the hot water we make, stop using most of the electric light, lanterns and candles that we use, and generally stop burning all of that stuff.

OR . . .

Find a way to burn it without producing CO₂ .

OR . . .

Find other ways to produce the energy we want WITHOUT burning that stuff.

THAT's why you're hearing so much these days about solar energy, wind energy, and nuclear energy.  Those don't produce ANY CO₂ .

(I didn't put bio-Diesel and Ethanol on the list. These are just methods to avoid making fuel from oil pumped out of the ground, but they produce just as much CO₂ when they burn.  So they're solutions to the problems and costs of oil production,  but they're no help at all with the CO₂ problem.)

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Technician A states that accidents avoidance is an additional feature on some electronic stability control systems. Technician B
Julli [10]

Answer:

Both A and B are right.

Explanation:

-Electronic Stability Control (ESC) helps drivers to avoid crashes by reducing the danger of skidding, or losing control as a result of over-steering. The ESC activates in the instant that a driver loses control.

-The Sway Control is an added feature that helps eliminate trailer sway and takes action when necessary via the electric brake system, to maintain road position.

5 0
3 years ago
What is the frequency of a radio wave with an energy of 3.686 × 10−24 j/photon?answer in units of hz?
Makovka662 [10]

To solve this problem, we have to use the formula:

E = h f

where E is total energy, h is Plancks constant 6.626x10^-34 J s, f is frequency

 

f = E / h

f = 3.686 × 10−24 J / (6.626x10^-34 J s)

<span>f = 5.56 x 10^9 Hz</span>

4 0
3 years ago
Generators use the magnetic field of which of the following to produce an electric current?
STatiana [176]
<span>Generators use the magnetic field of "Permanent Magnets"

In short, Your Answer would be Option D

Hope this helps!</span>
5 0
4 years ago
What is the element with the lowest electronegativity value?
Lerok [7]
<h2>Answer: Francium </h2>

Let's start by explaining that electronegativity is a term coined by Linus Pauling and is determined by the <em>ability of an atom of a certain element to attract electrons when chemically combined with another atom. </em>

So, the more electronegative an element is, the more electrons it will attract.

It should be noted that this value can not be measured directly by experiments, but it can be determined indirectly by means of calculations from other atomic or molecular properties of the element. That is why the scale created by Pauling is an arbitrary scale, where the maximum value of electronegativity is 4, assigned to Fluorine (F) and the <u>lowest is 0.7, assigned to Francium (Fr).</u>

7 0
4 years ago
Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103
son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

PE = -\frac{GMm}{R}

As M=m, then

PE = -\frac{Gm^2}{R}

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

U = -\frac{2Gm^2}{R}

dU = -\frac{2Gm^2}{R}

dKE = -dU

Therefore replacing we have that,

\frac{1}{2}mv^2 =\frac{Gm^2}{2R}

Re-arrange to find v,

v= \sqrt{\frac{Gm}{R}}

v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}

v = 816.7m/s

Therefore the  velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

2r = 2*10^3m

Therefore

dU = Gm^2(\frac{1}{R}-\frac{1}{2r})

v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}

v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}

v = 1.83*10^7m/s

Therefore the velocity when they are about to collide is 1.83*10^7m/s

7 0
3 years ago
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