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pav-90 [236]
4 years ago
9

The atmosphere of Mars is almost all carbon dioxide and the average surface pressure is 610 Pa (as compared with 101,000 Pa on E

arth). A model for the temperature in the atmosphere is given by: T = To (e)^-Cz, where C m/s?. where C=1.1 E-5 /m. Assume ideal gas behavior. The gravitational acceleration is g = 3.71 m/s2. Determine: (a) analytic equation for the variation of pressure with altitude (b) the altitude [m] where the pressure is 10 Pa.
Physics
1 answer:
Karolina [17]4 years ago
4 0

Answer:

   z = 3,737 10⁵ m

Explanation:

a) As they indicate that the atmosphere behaves like an ideal gas, we can use the equation

          P V = n R T

          P = (n r / V) T

We replace

         P = (n R / V) T₀ e^{- C z}

b) Let's apply this equation in the points

Lower

        .z = 0

         P₀ = 610 Pa

         P₀ = (nR / V) T₀

Higher.

         P = 10 Pa

          P = (n R / V) T₀ e^{- C z}

We replace

        P = P₀ e^{- C z}

        e^{- C z} = P / P₀

        C z = ln P₀ / P

        z = 1 / C ln P₀ / P

Let's calculate

        z = 1 / 1.1 10⁻⁵ ln (610/10)

        z = 3,737 10⁵ m

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Mechanical life support system

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4 0
3 years ago
Which type of wire would have the least resistance?
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The answer is E. Short and thick while cold.
4 0
4 years ago
what will be the focal lenght of a combined lens made by contact of two lenses of power +3D and -2D.​
4vir4ik [10]
  • P_1=+3D
  • P_2=-2D

\\ \bull\tt\dashrightarrow P=P_1+P_2

\\ \bull\tt\dashrightarrow P=+3D-2D=+1D

Now

\\ \bull\tt\dashrightarrow f=\dfrac{1}{P}

\\ \bull\tt\dashrightarrow f=\dfrac{1}{1}

\\ \bull\tt\dashrightarrow f=1m

3 0
2 years ago
With what tension must a rope with length 3.00 mm and mass 0.105 kgkg be stretched for transverse waves of frequency 40.0 HzHz t
VladimirAG [237]

Answer:

the tension of the rope is 34.95 N

Explanation:

Given;

length of the rope, L = 3 m

mass of the rope, m = 0.105 kg

frequency of the wave, f = 40 Hz

wavelength of the wave, λ = 0.79 m

Let the tension of the rope = T

The speed of the wave is given as;

v = f\lambda = \sqrt{\frac{T}{\mu} } \\\\where;\\\\\mu \ is \ mass \ per \ unit \ length\\\\\mu  = \frac{0.105}{3} = 0.035 \ kg/m\\\\v = f\lambda = 40 \times 0.79 = 31.6 \ m/s\\\\v =  \sqrt{\frac{T}{\mu} } \\\\v^2 = \frac{T}{\mu} \\\\T = v^2 \mu\\\\T = (31.6^2)(0.035)\\\\T = 34.95 \ N

Therefore, the tension of the rope is 34.95 N

4 0
3 years ago
The closest distance a book can be read from a pair of reading eyeglasses (Power = 1.55 dp) is 26.0 cm. What is the near distanc
Mnenie [13.5K]

Answer:

The image distance is 20.0 cm.

Explanation:

Given that,

Power = 1.55 dp

Distance between book to eye = 26.0+3.00=29.0 cm

We need to calculate the focal length

Using formula of focal length

f = \dfrac{1}{P}

Put the value into the formula

f=\dfrac{1}{1.55}

f=0.645\ m

f=64.5\ cm

We need to calculate the image distance

Using lens formula

\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}

\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{-u}

Put the value into the formula

\dfrac{1}{v}=\dfrac{1}{64.5}-\dfrac{1}{-29}

\dfrac{1}{v}=\dfrac{187}{3741}

v=20.0\ cm

Hence, The image distance is 20.0 cm.

5 0
3 years ago
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