Plane takes off, reaches mach 1.3, flies 1K miles, accelerates to mach 2.1, flies 100 mi. How many sonic booms?Posted on November 19th, 2010 No comments
First correct and complete answer was from @jchaager: A plane flying at mach 1 or faster produces one continuous sonic boom which “follows” the plane. At least two people said a supersonic plane makes 1 sonic boom without specifying that it was continuous. Not sure they got it or not.
An airplane, or any other object traveling at the speed of sound or faster produces a sonic boom continually until it decelerates below mach 1. There is a common misconception that a plane produces a singular sonic boom when it “breaks the sound barrier.” Not true. There is an almost equally common misconception that a plane produces another sonic boom at each whole multiple of speed of sound, as it “breaks the second sound barrier” at mach 2, and so on. Not true.
An object moving faster than the speed of sound produces a shock wave which is perceived by an observer as a “sonic boom” as it passes. An observer hears a sonic boom after the airplane passes, even though the plane may have “broken the sound barrier” hundreds of miles away.
One would have hoped that the famous “double sonic boom” of a Space Shuttle on final approach would cure the “sonic boom/sound barrier” myth. The double sonic boom of the Space Shuttle is caused by its curving approach. The shuttle passes the observers twice while still supersonic.
Here in Middle Georgia, sonic booms from U.S. Air Force fighters are a weekly occurrence, but I have a couple of noteable personal experiences with sonic booms.
One day (many years ago) while hunting, I heard sharp “crack” in the trees overhead. About two seconds later, I heard gunshot on next hill. A moment later, this was repeated: A “crack” overhead followed by a distant gunshot two seconds later. What was happening here was that another hunter a mile or so away was shooting at something (and missing), and the bullets were passing my position. Each bullet arrived at my location before the sound of the shot that fired it. As the supersonic bullet passed me, I heard its tiny sonic boom. I didn’t hear the gunshot until that sound had time to travel the mile or so from the shooter’s position to mine.
Now, these bullets “broke the sound barrier” within the barrel of the gun about a mile away from me. But I heard the sonic boom as the bullet passed me. It took a total of about five seconds for the sound of the gunshot – and any sound the bullet may have made by “breaking the sound barrier” – to reach me, but I heard the sonic boom from the bullet 20 feet over my head long before the sound of the gunshot reached me.
There is a very odd phenomenon that may be observed after a sonic boom if the airplane making it passed very close by. I observed this once when a U.S. Navy F-14 “buzzed” my ship while supersonic at mast-top level. The plane had approached from the port side while I was on the starboard side, so I never saw it coming. After the hellaceous blast of the sonic boom, I saw the plane departing very rapidly off the starboard side, and I heard the sound of its departing engines following it. But I also noticed what sounded like another plane’s engines departing rapidly off the port side of the ship, in the direction from which the F-14 had come.
What was happening here was that the sounds made by the plane were “catching up” to my position in reverse order. Sounds made 10 seconds before the plane passed us were arriving at the ship after the sounds made 9 seconds before the plane passed, and so on. This is a perfectly straightforward consequence of the supersonic flight of the plane, but it produces a very disconcerting effect. The observer hears the sounds of the departing plane in “normal” fashion, but the sounds the plane made as it approached are “played in reverse” as earlier sounds catch up to later sounds. It sounds like an invisible airplane flying away.
Ever since that day, whenever I see an airplane approaching but I can’t hear it, I plug my ears and cringe. I have never again heard a sonic boom from mast-top level, but I really, really don’t want to.
More about sonic booms: http://www.sciencetriviatweets.com/?p=1008
Posted on November 9th, 2010 No comments
There were several correct answers to yesterday’s question, but not all were as clear or complete as others. The first “complete” answer was from @jpwarren: Going to orbit of Mercury requires more fuel than leaving Solar System because must slow down. Quoting @jpwarren: “because you’re going deeper into the Sun’s gravity well and have to slow down.” Earlier correct but less clear/complete answers included:
- @jchaager: Entering orbit around mercury requires traveling faster than exiting the solar system due to the high gravity.
- @rozberk: I would say that it would take more power to fight increased gravitation forces than to exit from it.
And @rozberk had another answer about orbiting Mercury: “Because you’d just “have” to visit Vulcan while you’re there. :)”
As a spacecraft travels downsun, it gains kinetic energy (speed). To match Mercury’s orbital velocity, it must slow down tremendously. Thus, entering into orbit of Mercury with a starting point on Earth requires more rocket fuel than exiting the Solar System completely.
“Rocket fuel” is not necessarily the right term. The BepiColombo mission to Mercury will use solar ion drive to slow it down. This does not involve “rocket fuel” per se, but the spacecraft will still require a huge amount of propulsion to slow it down to match Mercury’s orbital velocity. The ion drive will do the job, but it will require a very long time to do so.
The current MESSENGER mission to Mercury uses conventional chemical thrusters, but it also makes extensive use of gravitational adjustments from planetary flybys to help it slow down to Mercury’s speed. That is why MESSENGER has already made three flybys of Earth since it was launched. Earth is helping to slow the spacecraft down (as are Venus, Mercury, and the Earth’s Moon).
BepiColombo is also making use of gravity assists from Earth, Venus, and Mercury to supplement the braking provided by its ion drive.
More about Mercury exploration: http://en.wikipedia.org/wiki/Exploration_of_Mercury and http://en.wikipedia.org/wiki/BepiColombo
Today’s Science Trivia: What chemical element that does not exist in nature is installed in most homes in many industrialized countries?Posted on September 27th, 2010 No comments
First correct answer was from @arachne182: Americium, specifically Am-241, is used in smoke detectors. This element does not exist naturally.
Many people answered with elements that don’t exist in elemental form in nature, such as aluminum or tungsten. Not what I was looking for, … We could argue about “what I was looking for” vs. “what the question says.” I think the question says what I meant it to say ;-}
Elements such as most metals do not exist in elemental form on earth because they combine with other elements (especially oxygen). But even if there is no naturally pure aluminum on earth, there is aluminum. And there probably is elemental aluminum in the universe outside Earth’s atmosphere. Probably a lot of it, in fact.
However, transuranic elements with atomic number > 92 exist only under conditions that humans create. They do not exist at all in nature. (Some sources hold that plutonium can exist in nature, but other sources traditionally consider it a synthetic element.)
Americium (atomic number 95) can only be created by neutron bombardment of plutonium or heavier elements. It is a synthetic element.
Most synthetic elements are so highly radioactive that they can only exist very briefly. Once created by neutron bombardment (and subsequent decay), they quickly decay into more stable forms. However, a few isotopes of some transuranic elements are stable enough to last for many years, even thousands of years. They are at least a little radioactive, but they stick around long enough for us to make some use of them.
Most known isotopes of Americium are too unstable to last long. Many are so radioactive that they are too dangerous to handle without special equipment. Am-241 is an exception.
Am-241 is a strong source of alpha radiation and emits very little of other types of radiation. Thus it is safe for use in homes.
This may sound absurd, unless you know a bit about the nature of ionizing radiation. A “pure” alpha source is not at all dangerous to humans, unless it is ingested or inhaled. Properly contained, it could be safely held in your bare hand. The alpha particles it emits can not penetrate the dead layer of skin on your body, so it can not harm living tissue beneath. Of course, there is no such thing as a “pure” alpha source. Even Am-241 emits some gamma radiation, but it emits little enough that a small amount of Am-241 is harmless.
Alpha radiation is very effective at ionizing air, but is completely blocked by the plastic housing of the smoke detector.
An ionizing smoke detector works by passing electric current through ionized air. Smoke blocking the radiation interrupts the current, so the detector knows there is a problem.
Other types of smoke detector do not use Americium or any other source of ionizing radiation. Each type has is advantages, but the ionizing type is very popular due to its low cost and reliability.
More about Americium in smoke detectors: http://en.wikipedia.org/wiki/Smoke_detector#Ionization
More about Am-241: http://en.wikipedia.org/wiki/Americium-241
Besides round, what shape of manholes prevents the cover from being accidentally dropped down the manhole?Posted on August 30th, 2010 No comments
Q: Why are manhole covers round? A: Because manholes are round.
There are actually two correct answers to this question. The first correct answer was from @mlv: A triangular manhole cover (specifically an equilateral triangle) can not fall down the hole. The first person with the other correct answer was @rozberk: A manhole cover in shape of a Reuleaux triangle (more generally, a Reuleaux polygon) also works.
A Reuleaux polygon is not a true polygon, as each side is curved. Each side is a segment of a circle. The overall shape is described as a “curve of constant width.” A Reuleaux polygon is formed by the intersection of circles of a given radius, each centered on the vertices of regular polygon with an odd number of sides. The simplest Reuleaux polygon is the Reuleaux triangle. It is close to the shape of a rotor of a Wankel rotary engine, but not quite. The rotor of a Wankel rotary engine has sides which are more flattened than those of a Reuleaux triangle.
So, picture an equilateral triangle. Draw a circle centered on each apex whose radius is equal to the length of a side of the triangle. The intersection of these three circles defines a Reuleaux triangle. Another way to look at it is to connect each pair of apices of the triangle with a segment of a circle whose center is at the opposite apex.
Now, imagine this shape rotating between two vertical lines which would move together or apart as needed to keep in contact with the rotating Reuleaux triangle. As apex A is in contact with the right vertical line, side B-C is in contact with the left vertical line. As the triangle rotates, the two vertical lines do not move together or apart, as the separation between them is a circle. When the triangle rotates so that apex B is in contact with the left vertical line, continued rotation places the right vertical line in contact with side A-C. Since side A-C is a segment of a circle centered on B, the two vertical lines remain exactly the same distance apart, and so on.
Thus, it is a “curve of constant width,” but not a circle.
Theoretically, if you increase the number of apices to infinity and the length of each side becomes infinitesimal, the Reuleaux polygon becomes a circle.
I am not aware of any actual manhole covers in the shape of a Reuleaux polygon.
A manhole cover shaped as an equilateral triangle also can not be dropped inside the manhole. Where I used to live in Nashua, New Hampshire, nearly half of the manholes are equilateral triangles. Quoting Wikipedia: “Nashua, New Hampshire may be unique in the U.S. for having triangular manhole covers …”
The manhole covers in Nashua are not Reuleaux triangles, but true equilateral triangles. This shape is not of constant width, but it can not be dropped inside the manhole.
More about manhole covers: http://en.wikipedia.org/wiki/Manhole_cover
And about Reuleaux polygons: http://en.wikipedia.org/wiki/Reuleaux_triangle
Posted on August 19th, 2010 No comments
Hint: Not “half-life.”
First correct answer was from @rozberk: The time required to reduce radioactive material by half in a living organism is “effective half-life.”
“Effective half-life” is usually much shorter than the radioactive material’s half-life. It is affected by both the physical half-life of the radioactive material and the organism’s purging mechanisms. So, while the organism is metabolizing/excreting the radioactive material, the radioactive material is decaying on its own.
Consider an example where a material is 100% absorbed by a living organism. Every atom is incorporated into the organism’s fat reserves or bones or whatever. In such a case, the effective half-life is equal to the half-life, because the only way the radioactive material is being reduced is by its own decay.
In a more realistic case, where some of the material is absorbed and some of it is removed from the body – whether exhaled as a gas or excreted as liquid or solid waste – the material would be removed from the body at a faster rate than either radioactive decay or metabolic processes alone would remove it.
More about effective half-life: http://en.wikipedia.org/wiki/Effective_half-life
Posted on July 13th, 2010 No comments
First correct answer is from @mlv: Geckos’ toes bend backward, allowing them to pull at the angle where the adhesion force is weakest.
You might compare the way a gecko lifts its feet to the way you might pull a boot from mud. A straight pull doesn’t work, but if you push sideways to let air under the boot, it comes out easily. Or if you try to pull a strong magnet straight off a metal surface, you can’t but if you rotate it to lift one edge first, you can lift it.
The adhesive force of a gecko’s feet is sometimes compared to static electricity, but it’s actually more like a chemical bond. It’s called van der Waals force.
A gecko has thousands of “hairs” on each toe, and each hair has hundreds of separate “spatulae” for a total of millions of spatulae. The total adhesion force is many times gecko’s own weight. A gecko could never overcome the adhesion of its toes with its own muscles if it could not bend its toes backward to peel them from the surface in the direction in which the force is weak.
More about geckos’ toes: http://en.wikipedia.org/wiki/Gecko#Gecko_toes:_setae_and_van_der_Waals_forces
Most of us have heard of “depleted uranium” used in armor-piercing projectiles, but what is depleted uranium?Posted on July 6th, 2010 1 comment
First correct answer was from @rozberk: “Depleted” uranium is the leftovers after uranium enrichment.
Depleted uranium is a byproduct of uranium enrichment. Natural uranium contains several isotopes. It is about 99.3% Uranium-238 and about 0.7% U-235. (Not exactly. There are small traces of other isotopes, too.) U-235 is the one that most easily undergoes nuclear fission in either a reactor or a bomb, so the enrichment process can produce uranium that is “enriched” in U-235. The resulting junk is almost pure U-238, which is no good for reactors or bombs.
Depleted uranium is still radioactive, but much less so than enriched, or even natural uranium. That’s because U-238 is more stable than U-235. In fact, depleted uranium is so dense and so low in radioactivity it is actually used as radiation shielding!
It is the density of depleted uranium that makes it so useful in armor-piercing munitions. I’m not sure why they don’t use lead, which is even denser than depleted uranium. Just guessing here, but maybe lead, even though more dense, is softer than depleted uranium, and therefore depleted uranium makes better armor-piercing shells. Maybe?
In either case, lead or depleted uranium, spraying it around the battlefield at 6,000 rounds per second is dubious environmental policy. Both are chemically toxic, and the uranium is slightly radioactive.
Some years ago, a few news reports stated that the “depleted uranium” being sprayed all over Iraq was what was left after using uranium in nuclear reactors. That was not true at all. What’s left over after uranium-bearing fuel rods are used up is “spent fuel,” and it’s NASTY! Spent nuclear fuel contains uranium-235 which did not get fissioned, a good deal of U-238, and a huge load of “fission fragments,” which are mostly highly radioactive isotopes which do not exist in nature.
In fact, this spent fuel is so radioactive and gives off so much heat that it can be used in long-term low-power generators, and has been used on some deep-space probes. In these cases, there is no active reactor controlling the power being produced, but the decay heat of all those radioactive atoms produces enough heat to power a small unmanned spacecraft for years.
It is true that something like “depleted uranium” can be recovered from spent nuclear fuel, but this is not widely done. It’s safer just to let the spent fuel decay a few decades in storage before trying to handle it in any complex way.
More about depleted uranium: http://en.wikipedia.org/wiki/Depleted_uranium
We hear that the light of distant galaxies is “redshifted,” but in photos, they don’t look red. Why not?Posted on June 17th, 2010 No comments
First correct answer was from @Alliterative: A distant galaxy’s visible light shifts toward the red end of the electromagnetic (EM) spectrum, but its ultraviolet shifts into the visible band, so the galaxy still looks white.
Astronomers say a galaxy “looks redder” as shorthand, but popular press does not clarify this shorthand sufficiently, in my opinion.
When an object is “redshifted,” what we can see shifted toward the red end of the EM specrum is its spectral lines. Its color doesn’t change.
A star (and therefore a galaxy, which is a large grouping of stars) emits electromagnetic energy in all parts of the spectrum. We only see a narrow part of this spectrum, which we call “visible light.” Our instruments can detect EM energy at levels below the energy of red (infrared) and abov the energy of violet (ultraviolet). That’s how we know that this energy outside the visible range exists.
Suppose, for simplicity’s sake, a distant galaxy emits only red, green, blue, and ultraviolet photons. If it were closer, it would appear white, as our eyes would see the red, green, and blue. We would not see the ultraviolet. Since a distant galaxy is receding from us, its red light is shifted into the infrared, its green light shifts to red, blue shifts to green, and ultraviolet shifts into the blue range. Thus, our eyes still see the galaxy a white, because energy is present in all bands of red, green, and blue.
When astronomers say that the galaxy’s light is shifted toward red, they are referring to the position of its “spectral lines.” These lines are dark bands in a star’s spectrum where the star’s light passes through a cooler medium (the star’s outer atmosphere) which absorbs only particular wavelenghts of light. As originally emitted by the plasma of the star, the light contained all possible energy levels. However, when it passes through a cool gas of intact atoms (mostly of hydrogen), the light is filtered as the atoms absorb particular wavelengths. Since the atoms of this cooler gas have electrons in discrete energy states, they can only absorb certain discrete wavelengths of light. The result is “holes” of darkness in the spectrum of light from the star.
When a star is not moving (much) in comparison to the Earth, we see these spectral lines in particular places in the spectrum, and each line can be associated with a particular element or compound. When the star is moving toward or away from the Earth, these lines appear “in the wrong place” in the spectrum. If the star is moving toward the Earth, we perceive its spectral lines as shifted toward the blue end of the EM spectrum. If the star is moving away from the Earth, as in a distant galaxy, we perceive its spectral lines as shifted toward the red end of the EM spectrum, or even beyond.
Hypothetically, visible light emitted by a star might be redshifted into radio waves, and x-ray energy could be redshifted into visible light.
In any case, when a star is redshifted, there is always plenty of EM energy beyond the visible range to be shifted into the range of visible light, so the star always appears pretty close to white.
More about redshift: http://en.wikipedia.org/wiki/Redshift
And more about spectral lines: http://en.wikipedia.org/wiki/Spectral_line
Posted on June 11th, 2010 No comments
Firs correct answer was from @mlv: The “waterhole” is a quiet band in the electromagnetic (EM) spectrum, corresponding to wavelengths between those emitted by hydroxyl ions and hydrogen atoms. Since hydroxyl and hydrogen combine to form water molecules, this quiet band was dubbed the “waterhole.”
Since the “waterhole” has little natural radio noise in it, compared to other parts of the EM spectrum, it’s a good set of wavelengths for interstellar communication. The Search for Extraterrestrial Intelligence concentrates on the “waterhole,” as any hypothetical extraterrestrial intelligence would understand its advantages, too.
More about the “waterhole”: http://en.wikipedia.org/wiki/Waterhole_(radio)
Term for the one-way flow of electrical current in a vacuum when negative charge applied to heated element.Posted on June 2nd, 2010 No comments
No correct answers to this one. The one-way flow of electrical current in a vacuum when the negative element is heated is “Edison effect.”
The term “thermionic emission” refers to the emission of ions (e.g. electrons) from a heated element, not to the flow of current. Edison effect current can not flow without thermionic emission, but the two phenomena are not the same thing. The difference between “thermionic emission” and “Edison effect” is like the difference between “gravity” and a waterfall. You can’t have the second without the first, but you can have the first without the second.
So, when a metal filament is heated in a vacuum and another metal element is inside the same chamber, electricity can flow through the vacuum. In such a vacuum tube, current flows only when the heated element is negative and the cooler element is positive. This is because the heat liberates electrons from the heated element (thermionic emission), and these electrons are then free to move through the vacuum when they are attracted to the positive element.
Thomas Edison discovered this phenomenon when investigating why his light bulbs burned out.
Sometimes the term “Edison effect” is applied to the thermionic emission of electrons from a heated metal, but this is incorrect.
A vacuum tube with one heated element and one cool one allows electric current to flow one way but not the other. It’s a diode. If another metal grid or filament is placed between the positive and negative elements, voltage on this control grid affects current flow. This is a triode, and it can act as an amplifier or an electronic switch.
More about Edison effect: http://en.wikipedia.org/wiki/Thermionic_emission