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VARVARA [1.3K]
3 years ago
9

How would improvement in use of renewable energy sources impact climate change sea-level rise?

Physics
1 answer:
bonufazy [111]3 years ago
7 0

Answer:

Almost immeasurably small.

Explanation:

The STORY is that humans are BAD for the environment and have caused a HUGE change in the amount of CO₂ in the atmosphere.

Let's look at the reports and draw our own conclusions.

Current CO₂ levels are 409.8 parts per million (PPM)

at the beginning of the Industrial revolution in the 1700's, the presumed beginning of the huge increase in CO₂ the level was about 280 PPM

For perspective lets assume we capture the whole atmosphere and squish it down to 2400 one liter bottles of air

That's 100 cases of 24 bottles per case.

We now separate all the air components into their own bottles

Nitrogen is 78% of our air, so we subtract 78 cases from our 100 leaving 22

Subtracting Oxygen at 21% of air leaves 1 case of liter bottles left

Of those 24 bottles, Argon makes up 0.93% of air so we subtract 22 bottles

The remaining two bottles contain all of the other gasses in our air, One of those bottles contains CO₂.

If we take the CO₂ levels from the 1700's at about 280 PPM as a baseline and assume ALL of the increase is human caused, that is (410 -280) / 280 = 46 % of the total.

The human caused addition of CO₂ would be 460 mililiters out of 2400 liters over the course of 250 years 

The claim is, that less than half of a liter of CO₂ out of 2400 liters of air is responsible for heating not only the gas in all the other bottles but also the surface of the earth itself.

Personally, it boggles my mind.

And it says NOTHING of a far more powerful greenhouse gas that is far more prevalent in the atmosphere...water vapor.

Water vapor is about 1% of air at sea level and about 0.4% overall. It was not considered in the above analysis because water vapor can condense out and is not a constant in the air.

Notice that there is about 100 times the amount of water vapor in the air as compared to CO₂. Water vapor also has between 4 and 8 times the greenhouse effect that CO₂ does.

Makes one wonder why we choose to pick on CO₂.

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If a car travels 60 mph for a distance of 180 miles, how much time<br> did it take?
jolli1 [7]

Answer:

3 hours

Explanation:

180 divided by 60 (mph means miles per hours by the way)

6 0
3 years ago
g An electron enters a region of space containing a uniform 1.63 × 10 − 5 T magnetic field. Its speed is 121 m/s and it enters p
kolbaska11 [484]

Answer:

i. The radius 'r' of the electron's path is 4.23 × 10^{-5} m.

ii. The frequency 'f' of the motion is 455.44 KHz.

Explanation:

The radius 'r' of the electron's path is called a gyroradius. Gyroradius is the radius of the circular motion of a charged particle in the presence of a uniform magnetic field.

                 r = \frac{mv}{qB}

Where: B is the strength magnetic field, q is the charge, v is its velocity and m is the mass of the particle.

From the question, B = 1.63 × 10^{-5}T, v = 121 m/s, Θ = 90^{0} (since it enters perpendicularly to the field), q = e  = 1.6 × 10^{-19}C and m = 9.11 × 10^{-31}Kg.

Thus,

         r = \frac{mv}{qB} ÷ sinΘ

But,  sinΘ =  sin 90^{0} = 1.

So that;

          r = \frac{mv}{qB}

            = (9.11 × 10^{-31} × 121) ÷ (1.6 × 10^{-19}  × 1.63 × 10^{-5})

            = 1.10231 × 10^{-28}   ÷ 2.608 × 10^{-24}

            = 4.2266 × 10^{-5}

            = 4.23 × 10^{-5} m

The radius 'r' of the electron's path is 4.23 × 10^{-5} m.

B. The frequency 'f' of the motion is called cyclotron frequency;

           f = \frac{qB}{2\pi m}

             =  (1.6 × 10^{-19}  × 1.63 × 10^{-5}) ÷ (2 ×\frac{22}{7} × 9.11 × 10^{-31})

             =  2.608 × 10^{-24} ÷  5.7263 × 10^{-30}

             = 455442.4323

          f  = 455.44 KHz

The frequency 'f' of the motion is 455.44 KHz.

3 0
4 years ago
Read 2 more answers
An object weighs 63.8 N in air. When it is suspended from a force scale and completely immersed in water the scale reads 16.8 N.
I am Lyosha [343]

Answer:

The density of this object is approximately 1.36\; {\rm kg \cdot L^{-1}}.

The density of the oil in this question is approximately 0.600\; {\rm kg \cdot L^{-1}}.

(Assumption: the gravitational field strength is g =9.806\; {\rm N \cdot kg^{-1}})

Explanation:

When the gravitational field strength is g, the weight (\text{weight}) of an object of mass m would be m\, g.

Conversely, if the weight of an object is (\text{weight}) in a gravitational field of strength g, the mass m of that object would be m = (\text{weight}) / g.

Assuming that g =9.806\; {\rm N \cdot kg^{-1}}. The mass of this 63.8\; {\rm N}-object would be:

\begin{aligned} \text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{63.8\; {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 6.506\; {\rm kg}\end{aligned}.

When an object is immersed in a liquid, the buoyancy force on that object would be equal to the weight of the liquid that was displaced. For instance, since the object in this question was fully immersed in water, the volume of water displaced would be equal to the volume of this object.

When this object was suspended in water, the buoyancy force on this object was (63.8\; {\rm N} - 16.8\; {\rm N}) = 47.0\; {\rm N}. Hence, the weight of water that this object displaced would be 47.0 \; {\rm N}.

The mass of water displaced would be:

\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{47.0\: {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 4.793\; {\rm kg}\end{aligned}.

The volume of that much water (which this object had displaced) would be:

\begin{aligned}\text{volume} &= \frac{\text{mass}}{\text{density}} \\ &\approx \frac{4.793\; {\rm kg}}{1.00\; {\rm kg \cdot L^{-1}}} \\ &\approx 4.793\; {\rm L}\end{aligned}.

Since this object was fully immersed in water, the volume of this object would be equal to the volume of water displaced. Hence, the volume of this object is approximately 4.793\; {\rm L}.

The mass of this object is 6.50\; {\rm kg}. Hence, the density of this object would be:

\begin{aligned} \text{density} &= \frac{\text{mass}}{\text{volume}} \\ &\approx \frac{6.506\; {\rm kg}}{4.793\; {\rm L}} \\ &\approx 1.36\; {\rm kg \cdot L^{-1}} \end{aligned}.

(Rounded to \text{$3$ sig. fig.})

Similarly, since this object was fully immersed in oil, the volume of oil displaced would be equal to the volume of this object: approximately 4.793\; {\rm L}.

The weight of oil displaced would be equal to the magnitude of the buoyancy force: 63.8\; {\rm N} - 35.6\; {\rm N} = 28.2\; {\rm N}.

The mass of that much oil would be:

\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{28.2\: {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 2.876\; {\rm kg}\end{aligned}.

Hence, the density of the oil in this question would be:

\begin{aligned} \text{density} &= \frac{\text{mass}}{\text{volume}} \\ &\approx \frac{2.876\; {\rm kg}}{4.793\; {\rm L}} \\ &\approx 0.600\; {\rm kg \cdot L^{-1}} \end{aligned}.

(Rounded to \text{$3$ sig. fig.})

7 0
2 years ago
This property of waves is the only property where the relationship between energy and this property are indirect or inverse
yaroslaw [1]

Answer: I don't understand

Explanation:

study and pay attention

4 0
2 years ago
How does the ratio of tin to copper affect the properties of the alloy bronze?
Pani-rosa [81]

Answer:

Stronger and harder than either of the pure metals

Explanation:

7 0
2 years ago
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