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fredd [130]
3 years ago
8

As the result of a thermal inversion the prevailing air temperature profile increases 1°C/100m above the ground level. To what m

aximum height will polluted air rise if it is released at ground level 15°C warmer than the atmosphere? The adiabatic lapse rate is −0.65°C/100m.
Physics
1 answer:
astra-53 [7]3 years ago
5 0

Answer:

909.1 m

Explanation:

Rate of temperature increase with 100 m elevation = 1°C

h = Maximum Height

Adiabatic lapse rate = -0.65°C/100 m

We have the relation

\dfrac{1^{\circ}C}{100}=\dfrac{15^{\circ}C}{h}+\left(-\frac{0.65}{100}\right)\\\Rightarrow \dfrac{1^{\circ}C}{100}+\left(-\frac{0.65}{100}\right)=\dfrac{15^{\circ}C}{h}\\\Rightarrow h=\dfrac{15^{\circ}C}{1.65^{\circ}C}\times 100\\\Rightarrow h=909.1\ m

The maximum height is 909.1 m

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A single-turn circular loop of radius 6 cm is to produce a field at its center that will just cancel the earth's field of magnit
djverab [1.8K]

Answer:

The current is  I  = 6.68 \  A

Explanation:

From the question we are told that  

     The radius of the loop is  r =  6 \ cm  = 0.06 \ m

     The  earth's magnetic field is B_e =  0.7G=  0.7  G * \frac{1*10^{-4} T}{1 G}  = 0.7 *10^{-4} T

      The  number of turns is  N  =1

Generally the magnetic field generated by the current in the loop is mathematically represented as

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Now for the earth's magnetic field to be canceled out the magnetic field generated by the loop must be equal to the magnetic field out the earth

         B  =  B_e

=>     B_e =  \frac{\mu_o  *  N  *  I  }{ 2 * r}

     Where  \mu is the permeability of free space with value  \mu _o  =   4\pi * 10^{-7} N/A^2

       0.7  *10^{-4}=  \frac{ 4\pi * 10^{-7}  * 1 * I}{2 * 0.06}

=>     I  =  \frac{2 *  0.06 *  0.7 *10^{-4}}{ 4\pi * 10^{-7} * 1}

       I  = 6.68 \  A

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