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sertanlavr [38]
2 years ago
7

Carbon dioxide and water molecules in the atmosphere will absorb:

Physics
1 answer:
alukav5142 [94]2 years ago
7 0

For CO2 we know that it has a useful life in the atmosphere of the earth, sometimes long. Carbon dioxide has the ability to absorb energy with a wavelength of 15 µm, this makes carbon dioxide a good absorbent of wavelengths that are in this range, such as infrared radiation for example .

At the same time the water vapor has the capacity to absorb the radiation in a 50 µm and 2.9 µm spectrum, infrared radiation fluctuates over the same range.

Therefore the correct answer is A.

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Is a tree a living or non living thing?
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Answer:

Living thing

Explanation:

Nonliving things do not move by themselves, grow, or reproduce.

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Which of the following is not an SI unit?
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Answer:

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Explanation:

SI Unit stands for the International Standard of Unit. Thus from the options given, metre is a measure of length which is an SI unit. Kilogram is a measure of mass of a goven matter which is also an SI unit. Second is a measure of time which is an SI unit. Pound is a measure of masswhich is not an SI unit  but has to be converted to kilograms which an SI unit.

8 0
3 years ago
One hundred turns of (insulated) copper wire are wrapped around a wooden cylindrical core of cross-sectional area 5.87 × 10-3 m2
Grace [21]

Answer:

The amount of charge that flow through a point is \Delta q  =  0.126 \ C

Explanation:

From the question we are told that

    The number of turns is  N  =  100

    The area is  A = 5.87 *10^{-3} m^2

       The total resistance is R = 14.9 \ \Omega

       The magnetic field in first direction  is  B_1 =  1.93 \ T

        The magnetic field in second   direction  is B_2 = -1.93 \ T

The change in magnetic field is evaluated as

     \Delta B  =  B_1 - B_2

substituting values

      \Delta B  =  1.93 - [- 1.93]

       \Delta B  =3.2 \ T

The induced emf due to the change is mathematically evaluated as

        e = NA \frac{\Delta B }{\Delta t }

This can also be mathematically represented as

     e = IR

     Where I can be mathematically represented as

     I =  \frac{\Delta q }{\Delta t}

So  

     e = \frac{\Delta q}{\Delta t } R

Now  

      \frac{\Delta q}{\Delta t } R    = NA \frac{\Delta B }{\Delta t }

=>   \Delta q  =  \frac{N A (\Delta B)}{R}

substituting values

     \Delta q  =  \frac{100 *  5.87 *10^{-3} (3.2}{14.9}

     \Delta q  =  0.126 \ C

     

   

     

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