All categories

3 years ago

Why can't passenger planes fly very high?

1 answer:

4 0

<span>It is not that jet engines perform better at higher altitude, but rather they are more fuel efficient. Performance is far greater at lower altitudes. Just think about when you are taking off and the airplane accelerates quickly and pitches way up. There's lots of extra thrust down low to allow that. The fuel/air ratio remains somewhat constant through the climb. As altitude increases, the air thin outs and therefore, so can the fuel flow. Airlines try for the most efficient routes and altitudes as possible to save money. They do however change altitudes in flight (higher or lower) when needed for weather and turbulence avoidance. --- And a note about the jet stream, it is relatively narrow and always curving, so the time an airliner would spend there is very short. And another thing, it flows mostly west to east isn't the U.S. so a westbound flight would be at a disadvantage. Airlines still fly high whether traveling East or West.</span>

You might be interested in

If the surface air pressure is 1000 mb and the pressure at the top of the atmosphere (75 km) is 0 mb, at what altitude would I f

Lana71 [14]

Answer: 5.5km

Explanation:

Atmospheric pressure will be 500 mb (that is half of the total 1000mb air pressure).

Pressure decreases with increasing altitude. This is because at At higher altitudes, there are fewer air molecules above a the known or given surface than a similar surface at lower levels.

Pressure decreasing with higher altitudes also means that air pressure decreases rapidly at lowerevels but more slowly at higher levels.

It is also known that more than half of the atmospheric molecules are located below 5.5 km(that is atmospheric pressure decreases within the lowest 5.5 km to about fifty(50) percent( that is 500 millibar).

8 0

3 years ago

Which of the following produces a physical change ?

stellarik [79]

Answer:

B: leaving metal outside in the rain until rust forms an it's surface

4 0

2 years ago

Which observational tool helped astronomers Arno Penzias and Robert Wilson discover the

iren2701 [21]

Answer:

A radio telescope helped the astronomers discover the CMB.

Explanation:

Penzias and Wilson while experimenting with a radio telescope in 1964, accidentally discovered the radiation that exists universally also known as the CMB.
This was used to support the "Big Bang Theory" and not the "Steady State Theory"
CMB is the faint cosmic radiation that fills up the universe. It provides important data for understanding early universe.
This data tells us about the composition of the universe and its age which raises new questions about the universe.

3 0

3 years ago

Group 7A of the periodic table contains the?

olga2289 [7]

The correct answer is A. Most reactive non metals.

Firstly if some one knows how to read the periodic table he would have no confusion in deciding whether group 7 has metals or non metals. Group 7 contains non metals so basically we can easily cancel out two options of metals.

Secondly group 7 non metals are the most reactive non metals as they need only one electron in order to complete their valence shell and become stable.

4 0

3 years ago

Read 2 more answers

Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface.

elena-s [515]

Answer:

$g=13.42\frac{m}{s^2}$

Explanation:

1) Notation and info given

$\rho_{center}=13000 \frac{kg}{m^3}$ represent the density at the center of the planet

$\rho_{surface}=2100 \frac{kg}{m^3}$ represent the densisty at the surface of the planet

r represent the radius

$r_{earth}=6.371x10^{6}m$ represent the radius of the Earth

2) Solution to the problem

So we can use a model to describe the density as function of the radius

$r=0, \rho(0)=\rho_{center}=13000 \frac{kg}{m^3}$

$r=6.371x10^{6}m, \rho(6.371x10^{6}m)=\rho_{surface}=2100 \frac{kg}{m^3}$

So we can create a linear model in the for y=b+mx, where the intercept b= $\rho_{center}=13000 \frac{kg}{m^3}$ and the slope would be given by $m=\frac{y_2-y_1}{x_2-x_1}=\frac{\rho_{surface}-\rho_{center}}{r_{earth}-0}$

So then our linear model would be

$\rho (r)=\rho_{center}+\frac{\rho_{surface}-\rho_{center}}{r_{earth}}r$

Since the goal for the problem is find the gravitational acceleration we need to begin finding the total mass of the planet, and for this we can use a finite element and spherical coordinates. The volume for the differential element would be $dV=r^2 sin\theta d\phi d\theta dr$ .

And the total mass would be given by the following integral

$M=\int \rho (r) dV$

Replacing dV we have the following result:

$M=\int_{0}^{2\pi}d\phi \int_{0}^{\pi}sin\theta d\theta \int_{0}^{r_{earth}}(r^2 \rho_{center}+\frac{\rho_{surface}-\rho_{center}}{r_{earth}}r)$

We can solve the integrals one by one and the final result would be the following

$M=4\pi(\frac{r^3_{earth}\rho_{center}}{3}+\frac{r^4_{earth}}{4} \frac{\rho_{surface}-\rho_{center}}{r_{earth}})$

Simplyfind this last expression we have:

$M=\frac{4\pi\rho_{center}r^3_{earth}}{3}+\pi r^3_{earth}(\rho_{surface}-\rho_{center})$

$M=\pi r^3_{earth}(\frac{4}{3}\rho_{center}+\rho_{surface}-\rho_{center})$

$M=\pi r^3_{earth}[\rho_{surface}+\frac{1}{3}\rho_{center}]$

And replacing the values we got:

$M=\pi (6.371x10^{6}m)^2(\frac{1}{3}13000 \frac{kg}{m^3}+2100 \frac{kg}{m^3})=8.204x10^{24}kg$

And now that for any shape the gravitational acceleration is given by:

$g=\frac{MG}{r^2_{earth}}=\frac{(6.67408x10^{-11}\frac{m^3}{kgs^2})*8.204x10^{24}kg}{(6371000m)^2}=13.48\frac{m}{s^2}$

4 0

3 years ago