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diamong [38]
3 years ago
6

Atmospheric pressure and density at the top of Mount Everest ​

Physics
1 answer:
Art [367]3 years ago
3 0

The summit of Mount Everest has an average pressure around 30 kPa. ... A barometer also measures variations in atmospheric pressure. As altitude increases, the air becomes thinner, the density of air decreases, and the pressure of the air decreases as well.

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A car traveling initially at 9.49 m/s accelerates at the rate of 0.988 m/s^2 for 3.05s. What is it’s velocity at the end of the
weeeeeb [17]

Answer:

12.50 m/s

Explanation:

Vi = 9.49 m/s

a = 0.988 m/s²

t = 3.05 s

Vf = ?

Vf = Vi + at

Vf = 9.49 + (0.988)(3.05)

Vf = 12.50 m/s

6 0
3 years ago
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If a rock sample has a mass of 1.17 g and a volume of 0.33 cm3, what type of rock is it?
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The correct answer is diamond

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kristine speeds past a parked police car at 32 m/s. The police car starts from rest with a uniform acceleration of 2.5 m/s^2. Ho
Digiron [165]

32 = 0 \times t +  \frac{1}{2} \times 2.5 \times t^{2}     \\  32 = 0 + 1.25 \times t {}^{2}  \\ 32 = 1.25t {}^{2}   \\   \frac{32}{1.25}  =  \frac{1.25t {}^{2} }{1.25}  \\ t {}^{2}  = 25.6 \\  \sqrt{t {}^{2} }  =  \sqrt{25.6}  \\ t = 5.1seconds \\

7 0
2 years ago
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
Lapatulllka [165]

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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3 years ago
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The answer to this is D
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2 years ago
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