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Artist 52 [7]
3 years ago
11

Increasing the concentration of greenhouse gases in Earth's atmosphere decreases the transparency of the atmosphere to infrared

light. decreases the transparency of the atmosphere to visible light. increases the transparency of the atmosphere to visible light. increases the transparency of the atmosphere to infrared light. Which of the following statements about global warming is NOT true? Computer models indicate that only human contributions to atmospheric carbon dioxide can account for recent global temperature rises Atmospheric carbon dioxide levels have risen dramatically over the past hundred years or so. Rises in global mean temperature tend to be closely associated with rises in atmospheric carbon dioxide concentrations. The global mean temperature in the modern era is the highest is has ever been
Physics
1 answer:
Artist 52 [7]3 years ago
5 0

Answer:

Decrease the transparency of the atmosphere

Explanation:

Different gases have characteristic absorption bands, so-called greenhouse gases have very high bands in the infrared region and small bands in the region of visible light.

   When the radiation  arrives to pass the greenhouse gases (CO²) and affects the earth's surface after several processes, the earth emits radiation in the infrared region, but this is absorbed by the gases, creating a net heating, since that the transparency of the atmosphere is  DECREASE for infrared radiation

Reviewing the correct answers is: Decrease the transparency of the atmosphere

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What is a microwave?
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an electromagnetic wave with a wavelength in the range 0.001–0.3 m, shorter than that of a normal radio wave but longer than those of infrared radiation. Microwaves are used in radar, in communications, and for heating in microwave ovens and in various industrial processes.

7 0
2 years ago
Three cars (car F, car G, and car H) are moving with the same velocity when the driver suddenly slams on the brakes, locking the
Nastasia [14]

Answer:

Car H

Explanation:

Frictional force is a resistant force. It is given as:

F = u*m*g

Where u = coefficient of friction

m = mass

g = acceleration due to gravity

From the formula above, we see that frictional force is dependent on the mass of object and the coefficient of friction.

Since they all have the same tires, the coefficient of friction between the tire and the floor is the same for each car. Acceleration due to gravity, g, is constant.

The only factor that determines the frictional force of each car is the mass. Hence, the more the mass, the more the frictional force.

So, the most massive car will have the most frictional force and hence, will come to a stop quicker than the others. The least massive car will have the least frictional force and so, will take a longer time to stop.

5 0
3 years ago
Two positive charges q1 = q2 = 2.0 μC are located at x = 0, y = 0.30 m and x = 0, y = -0.30 m, respectively. Third point charge
Wittaler [7]

Answer:

 F = 0.111015 N

Explanation:

For this exercise the force is given by Coulomb's law

        F = k q₁q₂ / r₂₁²

we calculate the electric force of the other two particles on the charge q1

Charges q₁ and q₂

the distance between them is

          r₁₂ = y₁ -y₂

          r₁₂ = 0.30 + 0.30

          r₁₂ = 0.60 m

let's calculate

          F₁₂ = 9 10⁹ 2 10⁻⁶ 2 10⁻⁶ / 0.60 2

          F₁₂ = 1 10⁻¹ N

directed towards the positive side of the y-axis

Charges 1 and 3

Let's find the distance using the Pythagorean Theorem

             r₁₃ = RA [(0.40-0) 2 + (0-0.30) 2]

             r₁₃ = 0.50 m

            F₁₃ = 9 10⁹ 2 10⁻⁶ 4 10⁻⁶ / 0.50²

            F₁₃ = 1.697 10⁻² N

The direction of this force is on the line that joins the two charges (1 and 3), let's use trigonometry to find the components of this force

           tan θ = y / x

           θ = tan⁻¹ y / x

          θ = tan⁻¹ 0.3 / 0.4

           tea = 36.87º

    The angle from the positive side of the x-axis is

         θ ’= 180 - θ

        θ ’= 180 - 36.87

        θ ’= 143.13º

       sin143.13 = F_13y / F₁₃

           F_13y = F₁₃ sin 143.13

           F{13y} = 1.697 10⁻² sin 143.13

           F_13y = 1.0183 10⁻² N

            cos 143.13 = F_13x / F₁₃

           F₁₃ₓ = F₁₃ cos 143.13

           F₁₃ₓ = 1.697 10⁻² cos 143.13

           F₁₃ₓ = -1.357 10-2 N

Now we can find the components of the resultant force

          Fx = F13x + F12x

          Fx = -1,357 10-2 +0

          Fx = -1.357 10-2 N

          Fy = F13y + F12y

         Fy = 1.0183 10-2 + ​​1 10-1

          Fy = 0.110183 N

We use the Pythagorean theorem to find the modulus

         F = Ra (Fx2 + Fy2)

         F = RA [(1.357 10-2) 2 + 0.110183 2]

         F = 0.111015 N

Let's use trigonometry for the angles

         tan tea = Fy / Fx

          tea = tan-1 (0.110183 / -0.01357)

          tea = 1,448 rad

to find the angle about the positive side of the + x axis

           tea '= pi - 1,448

           Tea = 1.6936 rad

6 0
3 years ago
the dimensions of a cuboidal block are 2.5 m * 2m * 1.2m its weight 900 n what is the minimum presseure exerted by the block on
aleksandr82 [10.1K]

Answer:

your answer is here 180

Explanation:

900/2.5 * 2 = 180

6 0
2 years ago
Read 2 more answers
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