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antiseptic1488 [7]
3 years ago
8

On the same spring day, a station near the equator has a surface temperature of 25°C, 15°C higher than the middle-latitude city

mentioned in part a. Vertical soundings reveal an environmental lapse rate of 6.5°C per kilometer and indicate that the tropopause is encountered at 16 km. What is the air temperature at the tropopause?
Physics
1 answer:
DedPeter [7]3 years ago
6 0

Answer:

The air temperature at the tropopause is - 79 °C

Explanation:

We know that a station near the equator has a surface temperature of 25°C

Vertical soundings reveal an environmental lapse rate of 6.5 °C per kilometer.

The tropopause is encountered at 16 km.

In order to find the air temperature at the tropopause we are going to deduce a linear function for the temperature at the tropopause.

This linear function will have the following structure :

f(x)=ax+b

Where ''a'' and ''b'' are real numbers.

Let's write T(x) to denote the temperature '' T '' in function of the distance

'' x '' ⇒

T(x)=ax+b

We can reorder the function as :

T(x)=b+ax (I)

Now, at the surface the value of ''x'' is 0 km and the temperature is 25°C so in the function (I) we write :

T(0)=25=b+a(0) ⇒ b=25 ⇒

T(x)=25+ax (II)

In (II) the value of ''a'' represents the change in temperature per kilometer.

Because the temperature decrease with the height this number will be negative and also a data from the question ⇒

T(x)=25-(6.5)x (III)

In (III) we deduced the linear equation. The last step is to replace by x=16 in (III) ⇒

T(16)=25-(6.5)(16)=-79

The air temperature at the tropopause is - 79 °C

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The Tevatron acceleator at the Fermi National Accelerator Laboratory (Fermilab) outside Chicago boosts protons to 1 TeV (1000 Ge
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a) v = c \cdot 0.04 = 1.2\cdot 10^{7} m/s

b) v = c \cdot 0.71 = 2.1\cdot 10^{8} m/s

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Here β = v/c, when v is the speed of the particle and c is the speed of light in vacuum.

Let's solve (1) for β.

\beta = \sqrt{1-\frac{1}{\left (\frac{KE}{mc^{2}}+1 \right )^{2}}}

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Now we can calculate the speed in each stage.

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\beta = \sqrt{1-\frac{1}{\left (\frac{0.75 MeV}{938.28 MeV}+1 \right )^{2}}}

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v = c \cdot 0.04 = 1.2\cdot 10^{7} m/s

b) Linac (400 MeV)

\beta = \sqrt{1-\frac{1}{\left (\frac{400 MeV}{938.28 MeV}+1 \right )^{2}}}

\beta = 0.71

v = c \cdot 0.71 = 2.1\cdot 10^{8} m/s

c) Booster (8 GeV)

\beta = \sqrt{1-\frac{1}{\left (\frac{8000 MeV}{938.28 MeV}+1 \right )^{2}}}

\beta = 0.994

v = c \cdot 0.994 = 2.97\cdot 10^{8} m/s

d) Main ring or injector (150 Gev)

\beta = \sqrt{1-\frac{1}{\left (\frac{150000 MeV}{938.28 MeV}+1 \right )^{2}}}

\beta = 0.999

v = c \cdot 0.999 = 2.997\cdot 10^{8} m/s

e) Tevatron (1 TeV)

\beta = \sqrt{1-\frac{1}{\left (\frac{1000000 MeV}{938.28 MeV}+1 \right )^{2}}}

\beta = 0.9999

v = c \cdot 0.9999 = 2.999\cdot 10^{8} m/s

Have a nice day!

4 0
4 years ago
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