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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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How far would a horse that rides at 5.5 m/s travel in 6.3 minutes?
larisa [96]

Answer: d = 2,079 m

Explanation:

5.5 m/s(6.3 min)(60 s/min) = 2,079 m

6 0
3 years ago
Waves diffract the most when their wavelength is?
Margarita [4]
<h2>Answer: about the same size of the gap </h2>

Diffraction happens when a wave (mechanical or electromagnetic wave, in fact, any wave) meets an obstacle or a slit .When this occurs, the wave bends around the corners of the obstacle or passes through the opening of the slit that acts as an obstacle, forming multiple patterns with the shape of the aperture of the slit.

Note that the principal condition for the occurrence of this phenomena is that <u>the obstacle must be comparable in size (similar size) to the size of the wavelength. </u>

<u></u>

In other words, <u>when the gap (or slit) size is larger than the wavelength</u>, the wave passes through the gap and does not spread out much on the other side, but when the gap size is equal to the wavelength, maximum diffraction occurs and the waves spread out greatly.

Therefore:

<h2>Waves diffract the most when their wavelength is <u>about the same size of the gap</u> </h2>

4 0
4 years ago
What would increase the force of gravity between two objects?
yulyashka [42]
By Newton's Law of Universal Gravitation.

F =  GMm/r²

Where F is Force of Gravitation, M = Mass of first object, m = mass of second object, r = distance of separation

From the formula, you can see that if the masses, M and m, increased, the value of F would definitely increase as well.

And if r increased the value of F would be reduced because you would be dividing by a bigger number, but when the value of r is decreased the value of F would be increased, because you would then be dividing by something smaller. Note the r is at the denominator of the formula.

So F would increase if there was increase in Masses and decrease in distance.

So the answer is C. a and b.
3 0
3 years ago
A rotating wheel requires 3.05-s to rotate through 37.0 revolutions. Its angular speed at the end of the 3.05-s interval is 97.9
Setler79 [48]

Answer:

\alpha=14.2rad/s^2

Explanation:

The formula that relates angular displacement with angular acceleration is:

\Delta \theta=\omega_i t+\frac{\alpha t^2}{2}

We can obtain \omega_i from the definition of angular acceleration:

\alpha=\frac{\Delta \omega}{\Delta t}=\frac{\omega_f-\omega_i}{t}

\omega_i=\omega_f-\alpha t

Putting all together:

\Delta \theta=(\omega_f-\alpha t) t+\frac{\alpha t^2}{2}=\omega_f t-\frac{\alpha t^2}{2}

Which, since we want the angular acceleration, is:

\alpha=\frac{2(\omega_f t-\Delta \theta)}{t^2}

And for our values is:

\alpha=\frac{2((97.9rad/s)(3.05s)-(37(2\pi rad)))}{(3.05s)^2}=14.2rad/s^2

5 0
4 years ago
Saturated steam at 125 kpa is compressed adiabatically in a centrifugal compressor to 700 kpa at the rate of 2.5 kg⋅s−1. the com
Tpy6a [65]
M° = 2.5 kg/sec
For saturated steam tables
at p₁ = 125Kpa
hg = h₁ = 2685.2 KJ/kg
SQ = s₁ = 7.2847 KJ/kg-k
for isotopic compression
S₁ = S₂ = 7.2847 KJ/kg-k
at 700Kpa steam with S = 7.2847
h₂ 3051.3 KJ/kg
Compressor efficiency
h =  0.78
0.78 = h₂ - h₁/h₂-h₁
0.78 = h₂-h₁ → 0.78 = 3051.3 - 2685.2/h₂ - 2685.2
h₂ = 3154.6KJ/kg
at 700Kpa with 3154.6 KJ/kg
enthalpy gives
entropy S₂ = 7.4586 KJ/kg-k
Work = m(h₂ - h₁) = 2.5(3154.6 - 2685.2
W = 1173.5KW
5 0
4 years ago
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