Re:Pioneers, NOT tourists!
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Space Tourism
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· Score: 1
I fear that someday we will reach a point where socioecononmic factors, public interest, and earth-based resources reach a point where large missions involving human launches would become an impossibility for public and private enterprise, left only to NASA or the ESA.
What do you mean, someday? This is the present reality.
How does Bob know Alice's address in the first place? Alice has to tell him. If Charlie is launching a man-in-the-middle attack, he can modify Alice's message, so that Bob opens up his firewall to Charlie's address instead.
It's a somewhat subtle point if you want to look at it that way. If we work in the zero momentum frame and assume that the emitted photons have no net momentum, then the rest mass of the product nucleus certainly has a smaller rest mass than the rest mass of the system before fusion (that's basically saying that the energy of the system minus the photons is less than the energy of the system!) But bear in mind that the rest mass of the initial system is not generally the sum of the rest masses of the fusing nuclei; their momentum make contributions too!
That sounds confusing, I know. That's because "mass-energy equivalence" is a pretty contorted way of viewing this process. The proper way to treat it is with four-vectors.
In terms of special relativity, proper mass is the length of the momentum four-vector. (And energy is the projection onto the time coordinate. That's why the two are equivalent when momentum is zero - the vector is perfectly aligned with the time coordinate, so its length would be the same as its projection.)
It's m_0 in the equation you supplied, but most people just use m.
During nuclear fission, the energy released comes from the nuclear binding energy. The binding energy of the nucleus before fission is lower (more negative) than the sum of the binding energies of the nuclei after fission. The difference is released as photons and kinetic energy.
"Mass-energy equivalence" isn't all it's cracked up to be, since "mass" in modern physics typically refers to proper mass, which isn't equivalent to energy. The famous equation E=mc^2 isn't actually all that useful, because it only holds true for systems with zero net momentum.
I've been running Mozilla from a non-world writable directory for ages. It lives in/usr/local/mozilla, which is owned by root. Unprivileged users can run it with no problem.
All you have to do is get root to run Mozilla *once* after installing it for the first time. Basically, the first time you launch it, Mozilla needs to generate some codepages and write into the mozilla directory, which only root can do.
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I fear that someday we will reach a point where socioecononmic factors, public interest, and earth-based resources reach a point where large missions involving human launches would become an impossibility for public and private enterprise, left only to NASA or the ESA.
What do you mean, someday? This is the present reality.
They can launch a man-in-the-middle attack on the phone line too, you know.
How does Bob know Alice's address in the first place? Alice has to tell him. If Charlie is launching a man-in-the-middle attack, he can modify Alice's message, so that Bob opens up his firewall to Charlie's address instead.
It's a somewhat subtle point if you want to look at it that way. If we work in the zero momentum frame and assume that the emitted photons have no net momentum, then the rest mass of the product nucleus certainly has a smaller rest mass than the rest mass of the system before fusion (that's basically saying that the energy of the system minus the photons is less than the energy of the system!) But bear in mind that the rest mass of the initial system is not generally the sum of the rest masses of the fusing nuclei; their momentum make contributions too! That sounds confusing, I know. That's because "mass-energy equivalence" is a pretty contorted way of viewing this process. The proper way to treat it is with four-vectors.
In terms of special relativity, proper mass is the length of the momentum four-vector. (And energy is the projection onto the time coordinate. That's why the two are equivalent when momentum is zero - the vector is perfectly aligned with the time coordinate, so its length would be the same as its projection.)
It's m_0 in the equation you supplied, but most people just use m.
Sure, except that equation doesn't display "mass-energy equivalence" since mass is no longer directly proportional to energy; momentum is involved.
That's why it's better to simply call energy energy, and (proper) mass mass. They're quite distinct concepts.
No.
During nuclear fission, the energy released comes from the nuclear binding energy. The binding energy of the nucleus before fission is lower (more negative) than the sum of the binding energies of the nuclei after fission. The difference is released as photons and kinetic energy.
"Mass-energy equivalence" isn't all it's cracked up to be, since "mass" in modern physics typically refers to proper mass, which isn't equivalent to energy. The famous equation E=mc^2 isn't actually all that useful, because it only holds true for systems with zero net momentum.
I've been running Mozilla from a non-world writable directory for ages. It lives in /usr/local/mozilla, which is owned by root. Unprivileged users can run it with no problem.
All you have to do is get root to run Mozilla *once* after installing it for the first time. Basically, the first time you launch it, Mozilla needs to generate some codepages and write into the mozilla directory, which only root can do.