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mario62 [17]
3 years ago
10

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

light. A. decreases the transparency of the atmosphere to visible light. B. increases the transparency of the atmosphere to visible light. C. increases the transparency of the atmosphere to infrared light.
Physics
1 answer:
kondaur [170]3 years ago
8 0

Answer:

Decreases the transparency of the atmosphere to infrared light.

Explanation:

When a large amount of green-house gases are present in the atmosphere, the layer of these gases become opaque to infrared radiation and radiation from the sun get trapped into these gases molecules. These excited molecules radiate this energy into our own atmosphere and that why the temperature of Earth is rising due to the Green-House effect.

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When a police officer is trying to decide if a driver is speeding, what is his point of reference. The speed limit
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2 years ago
State Archimedes' principle.
vampirchik [111]

Answer:

Archimedes' principle states that, when a body is partially or completely immersed in a fluid, it experiences an apparent loss in weight that is equal to the weight of the fluid displaced by the immersed part of the body.

Explanation:

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8 0
3 years ago
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A 1.0 kg piece of copper with a specific heat of cCu=390J/(kg⋅K) is placed in 1.0 kg of water with a specific heat of cw=4190J/(
Valentin [98]

Answer:

The temperature change of the copper is greater than the temperature change of the water.

Explanation:

deltaQ = mc(deltaT)

Where,

delta T = change in the temperature

m =mass

c = heat capacity

\frac{(deltaT)_{Cu}}{(deltaT)_{w}}=\frac{4190J/kg.K}{390J/kg.K}\\ \\(deltaT)_{Cu}=10.74(deltaT)_{w}

The temperature change in the copper is nearly 11 times the temperature change in the water.

So, the correct option is,

The temperature change of the copper is greater than the temperature change of the water.

Hope this helps!

7 0
3 years ago
The linear charge density on the inner conductor is and the linear charge density on the outer conductor is
Lubov Fominskaja [6]

Complete Question

The complete question is  shown on the first uploaded image (reference for Photobucket )

Answer:

The  electric field is  E = -1.3 *10^{-4} \ N/C

Explanation:

 From the question we are told that

    The linear charge density on the inner conductor is  \lambda _i  =  -26.8 nC/m  =  -26.8 *10^{-9} C/m

    The linear charge density on the outer conductor is

  \lambda_o  = -60.0 nC/m  =  -60.0 *10^{-9} \ C/m

     The position of interest is r =  37.3 mm =0.0373 m

Now this position we are considering is within the outer conductor so the electric field  at this point is due to the inner conductor (This is because the charges on the conductor a taken to be on the surface of the conductor according to Gauss Law )

Generally according to Gauss Law

         E (2 \pi r l) =  \frac{ \lambda_i }{\epsilon_o}

=>    E =  \frac{\lambda _i }{2 \pi *  \epsilon_o * r}

substituting values  

       E =  \frac{ -26 *10^{-9} }{2 * 3.142 *  8.85 *10^{-12} * 0.0373}

       E = -1.3 *10^{-4} \ N/C

The  negative  sign tell us that the direction of the electric field is radially inwards

    =>   |E| = 1.3 *10^{-4} \ N/C

5 0
3 years ago
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