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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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myrzilka [38]

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h = Vsinθ - gt
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Horizontal component x:
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In order for D (horizontal distance) to be maximum, dD/dθ = 0
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2V^2 cos2θ / g = 0
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Therefore it is now<span> shown that the maximum horizontal travelled is attained when the launch angle is π/4 radians, or 45°.</span>

6 0
3 years ago
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7 0
4 years ago
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v=\sqrt{\dfrac{k}{\rho}}

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v=\sqrt{\dfrac{40\times10^{6}}{10^3}}

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Using formula of wavelength

\lambda =\dfrac{v}{f}

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\lambda=\dfrac{200\times2\pi}{20\pi}

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Answer:

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Explanation:

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