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kakasveta [241]
3 years ago
11

In 1976, the SR-71A, flying at 20 km altitude (T = –56 0C), set the official jet-powered aircraft speed record of 3530 km/hr (21

93 mi/hr). Calculate the Mach number. Estimate the temperature at its front stagnation point.
Physics
1 answer:
Lapatulllka [165]3 years ago
6 0

To solve this problem we will apply the concepts related to the calculation of the speed of sound, the calculation of the Mach number and finally the calculation of the temperature at the front stagnation point. We will calculate the speed in international units as well as the temperature. With these values we will calculate the speed of the sound and the number of Mach. Finally we will calculate the temperature at the front stagnation point.

The altitude is,

z = 20km

And the velocity can be written as,

V = 3530km/h (\frac{1000m}{1km})(\frac{1h}{3600s})

V = 980.55m/s

From the properties of standard atmosphere at altitude z = 20km temperature is

T = 216.66K

k = 1.4

R = 287 J/kg

Velocity of sound at this altitude is

a = \sqrt{kRT}

a = \sqrt{(1.4)(287)(216.66)}

a = 295.049m/s

Then the Mach number

Ma = \frac{V}{a}

Ma = \frac{980.55}{296.049}

Ma = 3.312

So front stagnation temperature

T_0 = T(1+\frac{k-1}{2}Ma^2)

T_0 = (216.66)(1+\frac{1.4-1}{2}*3.312^2)

T_0 = 689.87K

Therefore the temperature at its front stagnation point is 689.87K

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iVinArrow [24]

To solve the problem we must know about the relationship between Speed, distance, and Time.

<h3>What is the relationship between Speed, distance, and Time?</h3>

We know that sped, distance, and time all are in a relationship to each other. this relationship can be given as,

\rm{Speed = \dfrac{Distance}{Time}

The speed of the light is 30,060.12 km/sec.

Given to us

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  • Light takes 499 seconds to reach earth from the sun.

We know that speed can be described as,

\rm{Speed = \dfrac{Distance}{Time}

Therefore,

<h3>What is the speed of the light?</h3>

\text{Speed of light} = \dfrac{\text{Distance between the earth and the sun}}{\text{Time taken by the light to travel the distance}}

Substitute the value,

\text{Speed of light} = \dfrac{15,000,000\ km}{499\ seconds}

\text{Speed of light} = 30,060.12\ km/sec

Hence, the speed of the light is  30,060.12 km/sec.

Learn more about Speed, distance, and Time:

brainly.com/question/15100898

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2 years ago
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A bicycle tire has a pressure of 7.00×105 N/m2 at a temperature of 18.0ºC and contains 2.00 L of gas. What will its pressure be
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Answer:

p_2 = 664081 N/m^{2}

Explanation:

from the ideal gas law we have

PV = mRT

P = \rho RT

\rho = \frac{P}{RT}

HERE  R is gas constant for dry air  =  287  J K^{-1} kg^{-1}

\rho = \frac{7.00 10^{5}}{287(18+273)}

\rho = 8.38 kg/m^{3}

We know by ideal gas law

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m_1 = \rho V_1 = 8.38 *2*10^{-3}

m_1 = 0.0167 kg

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m_2 = \rho V_i - V_removed

m_2 = 8.38*(.002 - 10^{-4})

m_2 = 0.0159 kg

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V, R and T are constant therefore we have

P is directly proportional to mass

\frac{p_2}{p_1}=\frac{m_2}{m_1}

p_2 = p_1 * \frac{m_2}{m_1}

p_2 =7*10^{5} * \frac {.0159}{0.0167}

p_2 = 664081 N/m^{2}

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Answer:

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Explanation:

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