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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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air mass

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1-Autotrophic plants are also called
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When atoms bond with each other, they exchange<br> A.) Protons<br> B) Neutrons <br> C)Electrons
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C)Electrons................................
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3 years ago
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6 0
3 years ago
A 200 kg weather rocket is loaded with 100 kg of fuel and fired straight up. It accelerates upward at 35 m/s2 for 35 s , then ru
r-ruslan [8.4K]

Answer:

T = 295.57 s

Explanation:

given,

mass of the rocket = 200 Kg

mass of the fuel = 100 Kg

acceleration = 35 m/s²

time, t = 35 s

time taken by the rocket to hit the ground, = ?

Final speed of the rocket when fuel is empty

using equation of motion

v = u + a t

v = 0 + 35 x 35

v = 1225 m/s

height of the rocket where fuel is empty

v² = u² + 2 a s

1225² = 0 + 2 x 35 x h₁

h₁ = 21437.5 m

After 35 s the rocket will be moving upward till the final velocity becomes zero.

Now, using equation of motion to find the height after 35 s

v² = u² + 2 g h₂

0² = 1225² + 2 x (-9.8) h₂

h₂ = 76562.5 m

total height = h₁ + h₂

          = 76562.5 m + 21437.5 m = 98000 m

now, time taken by before the rocket hit the ground

using equation of motion

s = u t +\dfrac{1}{2}at^2

-13500 = 1225 t -\dfrac{1}{2}\times 9.8 \times t^2

negative sign is used because the distance travel by the rocket is downward.

4.9 t² - 1225 t - 13500 = 0

t = \dfrac{-(-1225)\pm \sqrt{1225^2 - 4\times 4.9 \times (-13500)}}{2\times 4.9}

t = 260.57 s

neglecting the negative sign

total time the rocket was in air

T = t₁ + t₂

T = 35 + 260.57

T = 295.57 s

Time for which rocket was in air is equal to 295.57 s.

6 0
3 years ago
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