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uranmaximum [27]
3 years ago
5

Devise an exponential decay function that fits the given​ data, then answer the accompanying questions. Be sure to identify the

reference point ​(tequals ​0) and units of t. The pressure of a certain​ planet's atmosphere at sea level is approximately 800 millibars and decreases exponentially with elevation. At an elevation of 35 comma 000 ​ft, the pressure is​ one-third of the​ sea-level pressure. At what elevation is the pressure half of the​ sea-level pressure? At what elevation is it 2 ​% of the​ sea-level pressure?
Physics
1 answer:
7nadin3 [17]3 years ago
6 0

Answer:

22145.27733 ft

124984.76055 ft

Explanation:

The equation of pressure is

P=P_0e^{-kh}

where,

P_0 =Atmospheric pressure = 800 mbar

k = Constant

h = Altitude = 35000 ft

P=\dfrac{1}{3}P_0

\dfrac{1}{3}P_0=P_0e^{-k35000}\\\Rightarrow \dfrac{1}{3}=e^{-k35000}\\\Rightarrow 3=e^{k35000}\\\Rightarrow ln3=k35000\\\Rightarrow k=\dfrac{ln3}{35000}\\\Rightarrow k=3.13\times 10^{-5}

Now

P=\dfrac{1}{2}P_0

ln2=kh\\\Rightarrow h=\dfrac{ln2}{k}\\\Rightarrow h=\dfrac{ln2}{3.13\times 10^{-5}}\\\Rightarrow h=22145.27733\ ft

The altitude will be 22145.27733 ft

P=0.02P_0

0.02P_0=P_0e^{-kh}\\\Rightarrow 0.02=e^{-3.13\times 10^{-5}h}\\\Rightarrow ln0.02=-3.13\times 10^{-5}h\\\Rightarrow h=\dfrac{ln0.02}{-3.13\times 10^{-5}}\\\Rightarrow h=124984.76055\ ft

The elevation is 124984.76055 ft

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3 0
2 years ago
4. The Mariana trench is in the Pacific Ocean and has a depth of approximately 11,000 m. The density of seawater is approximatel
Alex73 [517]

Explanation:

It is known that relation between pressure and density is as follows.

            P = \rho gh

where,    P = pressure

     \rho = density

            g = acceleration due to gravity

            h = height

Putting the given values into the above formula as follows.

              P = \rho gh

                 = 1025 \times 9.8 \times 11000

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Now, relation between pressure and force is as follows.

                P = \frac{F}{A}

or,            F = PA

                F = 110495000 \times \pi \times (0.1)^{2}

                   = 3.47 \times 10^{6} N

Thus, we can conclude that a force of 3.47 \times 10^{6} N can be  experienced at such depth.

3 0
2 years ago
A cubical block of iron 10 cm on each side is floating on mercury in a vessel. (i) What is the height of the block above the mer
Gre4nikov [31]

Answer:

i 5.3 cm ii. 72 cm

Explanation:

i

We know upthrust on iron = weight of mercury displaced

To balance, the weight of iron = weight of mercury displaced . So

ρ₁V₁g = ρ₂V₂g

ρ₁V₁ = ρ₂V₂ where ρ₁ = density of iron = 7.2 g/cm³ and V₁ = volume of iron = 10³ cm³ and ρ₂ = density of mercury = 13.6 g/cm³ and V₂ = volume of mercury displaced = ?

V₂ = ρ₁V₁/ρ₂ = 7.2 g/cm³ × 10³ cm³/13.6 g/cm³ = 529.4 cm³

So, the height of iron above the mercury is h = V₂/area of base iron block

= 529.4 cm³/10² cm² = 5.294 cm ≅ 5.3 cm

ρ₁V₁g = ρ₂V₂g

ii

ρ₁V₁ = ρ₃V₃ where ρ₁ = density of iron = 7.2 g/cm³ and V₁ = volume of iron = 10³ cm³ and ρ₃ = density of water = 1 g/cm³ and V₃ = volume of water displaced = ?

V₃ = ρ₁V₁/ρ₃ = 7.2 g/cm³ × 10³ cm³/1 g/cm³ =  7200 cm³

So, the height of column of water is h = V₃/area of base iron block

= 7200 cm³/10² cm² = 72 cm

7 0
3 years ago
Is a force necessary to keep an object in motion?
weqwewe [10]
Well according to Newton’s first law of motion, a body will remain in the state of rest or linear motion provided that an *external force* has been applied. So no, a force doesn’t need to keep a body to remain in linear motion, because F=ma, during uniform linear motion velocity is constant, hence acceleration is zero, so F=0
5 0
2 years ago
The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz. Find the possible range of wavelengths in ai
taurus [48]

Answer:

The possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.

Explanation:

Given that,

The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz.

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To find,

The wavelength range for the corresponding frequency.

Solution,

The speed of sound is given by the following relation as :

v=f_1\lambda_1

Wavelength for f = 45 Hz is,

\lambda_1=\dfrac{v}{f_1}

\lambda_1=\dfrac{343}{45}=7.62\ m

Wavelength for f = 375 Hz is,

\lambda_2=\dfrac{v}{f_2}

\lambda_2=\dfrac{343}{375}=0.914\ m/s

So, the possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.

6 0
3 years ago
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