How long did it take to cut the lino?

It looks like the site is under maintenance. I'd give it a couple days and try again....

Sideways isn't the same thing. Newton's first law says the water will slosh out and the object will try to stay still relative to 'the universe' as the glass moves. If you move the glass up, however, the buoyant forces will increase. If the object if floating, it will continue to do so.

So do your experiment accelerating upwards. Of course it will be a mess when you stop :-)

I assumed the OC meant the person was floating in the water, so buoyant forces keep it at the top. When you accelerate, the person's weight increases, but the buoyant force increases exactly the same amount, so everything would stay where it was.

If it were a penny, then you are right - it would already be on the bottom and would stay there.

Agreed. Body won't hit the bottom of the tank, but the acceleration on the body will be identical. Being submerged avoids impact concentrated pressure on contact points (as you'd get from being in a seat) but acceleration on the body as a whole is identical.

To add more fun, if your foot is about a foot long and about 3 inches wide and a couple tall, you have about a square foot of skin, so there would be 144 * 260 psi = 37,440 lbs (or ~19 tons) of force squeezing the contents of your foot up your leg into your head like toothpaste.

The pressure gradient is the problem. If you were 'vertical' (head in direction of acceleration) the pressure on your toes and legs at 100G would be, let's see: 62.4 lbs/ft^2 / 144 in^2/ft * 100G * 6 ft = 260 PSI while your head would be 62.4/144100.5 = 21 PSI. So basically your legs will be squeezed into your chest and head. Not good. (Sorry for the stupid Imperial units...)

The person displaces their weight in water - that's the "on Earth" definition. The more general definition is they displace their mass. Their weight is m * a and the water they displace is rho * V * a (rho is density, V is volume, and a is acceleration). Where you float is therefore

m * a = rho * V * a

So you cancel the a on both sides and get

m = rho * V

(for fun note that rho * V is the mass of the water). So the general form of buoyancy is that you float where you displace your mass in water. Note this is completely independent on acceleration (g's). So the person will stay in the same position, unless compressed by the extra pressure of the pool, at which point they will "sink" (move in the direction of the acceleration vector). Of course getting compressed is the "then you die" part of the problem...

Don't think so. Position in the tank is determined by the person's buoyancy and the water pressure gradient, and both of those are determined by gravity on earth, and acceleration in space. The person should stay at the same position

I respectfully disagree. If the tank were upright on the ground, and a person got in, they would float at a certain depth, as their buoyant force (which comes from water pressure which depends on gravity) matches the weight of the person (which is also dependent on gravity). When you accelerate, the person's weight goes up, and the water pressure (and bouyant force) go up exactly the same amount, so they would stay at the same depth. The problem is the water pressure gradient at high Gs would kill them.

Edit: Spelling

I don't think this helps, since at high accelerations the pressure gradient in the water will probably kill you. (See longer answer below)

The problem is that the pressure experienced under water actually comes from gravity (acceleration). If you accelerate at 10G the pressure versus depth under water goes up 10x (relative to Earth). So being a meter under water is the same as being 10 m under water on Earth. At 100G, it goes up to 100m equivalent on Earth, which would squish you. Also, if you were upright, head just below the surface, your feet would be at 250 meters deep equivalent and your head at 2 meters equivalent, so it would squish all your feet and legs into your middle/head. It would not only kill you, but would be really awful to look at...

No. The fuel pellet is inside the hohlraum. Details and image here: https://lasers.llnl.gov/news/frustraum-hohlraum-design-is-shaping-up

Yes! I was lucky enough to get a tour of the NIF, and I asked the guide what the conversion efficiency was of the UV conversion (as those can be pretty inefficient), and he said 50%. Think about that - you go through the whole multi-building-sized process of building hundreds of terawatts of laser beams, and the last step - like 25 feet from the target it looked like - you lose HALF of the energy to UV-upconversion. Then 90% of the X-ray energy is lost. So you've got to start with a LOT of laser power. See https://lasers.llnl.gov/about/how-nif-works/final-optics.

Here's a video - the very end shows the lasers hitting the walls of the "package" holding the sphere with the fuel inside. The lasers hitting the walls raises their temperature so they produce X-rays. The X-rays ablate the outside of the fuel sphere, which flies outward. The inside of the sphere reacts to the ablation by blowing inwards, compressing the fuel.

If you do a couple google searches I'm sure you'll find an appropriate level explanation.

I believe you mean "specular" for "spectral" in the above.

Spectral measurements provide light intensity vs wavelength (or 'color') while specular measurements are of color including surface 'roughness'.

Beyond Hunter labs, a great read for beginner or experienced person is https://en.wikipedia.org/wiki/CIE_1931_color_space

The "UFO" is lens flare from the street lamp in the lower right.

(Assuming one is unfamiliar, lens flares are image artifacts from the lens and sensor. In this case, a surface - probably the back of the lens - is reflecting light bouncing off the sensor back into the sensor. These are easy to identify: Draw a line from the "UFO" through the middle of the image. You will find a bright light source at a point equidistant from the middle of the image, in this case the street light.)

As a matter of fact, if you look to the upper left of the "UFO" in the sky, you can see two faint "UFO"s which are the lower two lights by the same light pole.