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yan [13]
3 years ago
7

Cutting down the forest reduces the size of animal habitats, increases erosion, and results in an increase in the amount of carb

on dioxide in the atmosphere. In order to stop the harmful effects of cutting down trees, humans can A. clean up litter along the sides of the road. B. clear away the tree trunks and leave the land bare. C. dig up the land and use it as a landfill. D. plant more trees where forests are cut down.
Physics
1 answer:
Julli [10]3 years ago
6 0

Answer: D. plant more trees where forests are cut down.

Explanation:

Deforestation has to do with the cutting down of trees. Cutting down the forest reduces the size of animal habitats, increases erosion, and results in an increase in the amount of carbon dioxide in the atmosphere.

Ok order to tackle the above, it is necessary that we should plant more trees where forests are cut down.

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Within a period of the periodic table, how do the properties of the elements vary?
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Elements in the same period have the same number of electron shells; moving across a period (so progressing from group to group), elements gain electrons and protons and become less metallic. This arrangement reflects the periodic recurrence of similar properties as the atomic number increases.

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For heat transfer purposes, an egg can be considered to be a 5.5-cm-diameter sphere having the properties of water. An egg that
Harrizon [31]

Answer:

The time taken is   t = 40007 sec  

Explanation:

From the question we are told that

   The diameter of the egg is d_e = 5.5cm  = \frac{5.5}{100} = 5.5*10^{-2}m

    The initial temperature of egg the T_e = 4.3^{o}C

     The temperature of the boiling water T_b = 100^oC

    The heat transfer coefficient is  H  = 800 W/m^2 \cdot K

    The  final temperature is T_e_f = 74^oC

     The  thermal  conductivity of water is k = 0.607 W/m^oC

     The diffusivity of the egg \alpha = 0.146 * 10^{-6} m^2 /s

Using one term approximation

We have the

            \frac{T_e_f - T_b}{T_e - T_b}  = Ae^{-\lambda ^2 \tau}

The radius is  r = \frac{5.5*10^ {-2}}{2} =2.75*10^{-2}m     Note that this radius is approximation to that of  a real egg

    Now we need to obtain the Biot number which help indicate the value of A  \ and \ \lambda to use in the above equation

     The Biot number is mathematically represented as

               Bi = \frac{H r}{k}

Substituting values  

               Bi = \frac{800 * 2.75 *10^{-2}}{0.607}

                    = 36.24

So for this value  which greater than 0.1 the  coefficient \lambda_1 \ and  \ A_1 is  

        \lambda = 3.06632

        A = 1.9942

Substituting this into equation 1 we have

          \frac{74- 100}{4.3 - 100} = 1.9942 e^{-(3.0632^2) \tau}

          0.2717= 1.9942 e^{-(3.0632^2) \tau}

          0.2717= 1.9942 e^{-9.383 \tau}

           0.13624 =  e^{-9.383 \tau}

Taking natural log of both sides

           -1.993 =  -9.383\  \tau

          \tau =  0.2124

    The time required for the egg to be cooked is  mathematically represented as

          t = \frac{\tau r^2}{\alpha }

substituting value  is  

         = \frac{0.2124 * 2.75 *10^{-2}}{0.146 *10^{-6}}

         t = 40007 sec  

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I just had this question, the awnser is A.

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