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29

When you would enter the water, you need to "get the water out of the way". Say you need to get 50 liters of water out of the way. In a very short time you need to move this water by a few centimeters. That means the water needs to be accelerated in this short time first, and accelerating 50 kg of matter with your own body in this very short time will deform ...


18

Taking your question literally, you can see a single barium ion: The TRI┬ÁP group has achieved capturing a single barium ion in a Paul trap. The images show Coulomb crystals formed by a decreasing number of laser-cooled ions as detected with an EMCCD camera. This forms an important step towards the planned experiments on single radium ions to measure ...


14

The tension adds stress which brings the cable closer to the breaking point. With the additional stress of the rotor blades impacting it can reach the breaking point easier. In the case of string cut with scissors the tension on the string helps prevent the string from riding in between the blades separating them.


11

The ocean surface is not as hard as the ground but if you drop from a plane, you would hit it with such a high velocity that the pressure would most likely kill you or cause very serious damage. Considering air resistance, the terminal velocity of a human, right before reaching the water, would be at most some $150\text{ m/s}$. If you weigh $70\text{ kg}$, ...


11

Let's look at this another way: you're just moving from one fluid to another. Sounds harmless, right? By specification of the problem, we're at terminal velocity when we hit the water. The force of drag (in both mediums) is roughly: $$ F_D\, =\, \tfrac12\, \rho\, v^2\, C_D\, A = \rho \left( \frac{1}{2} v^2 C_D A \right) $$ You can imagine that ...


11

Consider jumping into a swimming pool. Do a barrel-roll (sorry I mean cannon ball, that just kind of slipped out). It's fun, you enter the water nicely and make a huge splash, probably soaking your sister in the process (that'll learn her). Now do a belly flop. Not as fun. You displace exactly the same amount of water in the same time, but this time there is ...


9

Materials snap when their fracture stress is exceeded. When a material is under tension, it has a stress applied to it. The higher the tension, the higher the stress. So a very tight cable is already close to it's fracture limit and any additional stress may exceed the limit. When it's loose, the additional stress may not be enough to exceed the fracture ...


8

You are indeed correct that there are so far no direct detections of gravitational waves, and no serious scientists are claiming (apart from excusable moments of being carried away) that the BICEP result constitutes such a direct detection. However, there is absolutely no evidence that gravitational waves don't exist, and plenty of indirect evidence that ...


7

Surface coating of an integrating sphere is optimized for low losses. This white coating (barium sulfate or PTFE) acts like an ideal lambertian scatterer. all light is scattered (Ok, not 100%, but a very high percentage like 99,5%. See ressources) it is emitted in the hemisphere following the cosine law: perpendicular to the surface it's highest. ...


6

I'm not a physicist. So I am treading very carefully attempting to answer a question here... :) A physical example that may help explain this is rock skipping. When you skip a rock, it will 'bounce' off of the water when at high speeds. Eventually it slows enough to no longer bounce but 'sink' into the water. Picture your body doing the same thing. Your ...


4

When a string it cut, it is not fracturing. There is a shear force applied to the surface of part of the string that interrupts the fibers by cutting. If you had a perfect pair of scissors where the blades had 0 gap even under load (e.g. trying to cut the string), then the tension on the string probably has little bearing. One thing that could result in ...


4

"White" light is light of multiple frequencies, usually a broad band of different frequencies ranging from ~($390nm$ to $700nm$) though white on your computer monitor consists of only three different frequencies (red green and blue). It is possible to send multiple frequencies down an optical fiber. However the fiber will act like a prism and diffract the ...


4

The way your question is phrased, it looks like you are expecting the mass to change. In that case the only change will be a slight mass loss due to evaporation, but the rate of evaporation is a variable - dependant on room temperature, air pressure humidity and how still the air is above the sample.


4

First, to be clear on what the graph is showing: as a function of the possible mass of the Higgs, it plots the fraction of Higgs bosons that will decay via each individual channel. Before we knew the mass of the Higgs boson, a plot like this one was useful for identifying the best channels to look at to detect the Higgs in various mass ranges. For example, ...


4

If the 10 cm was a measured value with a corresponding error estimate, then the results would be different. Suppose the bar was $10 \text{ cm} \pm 0.10 \text{ cm}$. Adding exactly $1 \text{ cm}$ to the length of the bar would yield $11 \text{ cm} \pm 0.10 \text{ cm}$, whereas increasing the length of the bar by 10% yields $11 \text{ cm} \pm 0.11 \text{ cm}$. ...


3

For aluminum 3003F between 4K and 300K see: http://www.cryogenics.nist.gov/MPropsMAY/3003F%20Aluminum/3003FAluminum_rev.htm For alumimum 5083 between 4K and 300K see: http://www.cryogenics.nist.gov/MPropsMAY/5083%20Aluminum/5083Aluminum_rev.htm for aluminum 6061-T6 between 4K and 300K see: ...


3

$10\%$ of $10\text{ cm}$ is $1\text{ cm}$. $10\text{ cm}\times\frac{10}{100}=1\text{ cm}$. I really don't see how this is a 'nit-pick', or 'relative' vs 'absolute', as suggested in the comments. Are you absolutely, $100\%$, or at least relatively, sure you got the quote right? (There are very close variations of it, where indeed it would be rather stupid ...


2

It sounds as though you might not have "filled" the lens properly. An infinite conjugate microscope objective, for example, has a specified numerical aperture when it is driven by for a collimated beam of a specified beamwidth and apodisation (i.e. whether Gaussian, uniform or so forth). See my drawing below: whence you can understand the fundamental ...


2

We have only looked back to the CMB time thus far - i.e. about 300,000 years after BB. Now having viewed this gravitational imprint on the CMB, we have pushed back our earliest observation to about 10^-35 seconds. That's a huge leap in the scope of our comprehension (a factor of about 10^47), and it's that which is probably the most impressive aspect of this ...


2

You are right when you say that one peak is due to $\mathrm{OH}$ groups and the other one to $\mathrm{CH_2}$ groups. The protons in each groups are chemically equivalent and contribute to the same peak. You should compute the area below each peak. Since there are 4 $\mathrm{CH_2}$ protons and only 2 $\mathrm{OH}$ protons, my guess is that one peak's area ...


2

All these answers are correct and describe reasons for the easier breaking of tightened string, but there's one more which contributes and arises from purely geometric considerations : consider a string from A to B and a force F applied in the middle. If the string is loose, you get this: If it is tight, that:


2

This is a link with the conclusions from the experiment and which observations lead to which conclusions. The Conclusion When Rutherford mathematically investigated the results he proposed a model that explained the results that Geiger and Marsden obtained. The fact that the vast majority of the alpha particles got straight through led Rutherford to ...


2

One thing would be that it's not practical to use '%' in that example. I can easily imagine metal worker going nuts if whole 'tech spec' is in ratios. Aftervards you can't just decrease by 10% and get the same original length. Or maybe because he didn't specify that it's 10% of original length. I have this 'dejavu' feeling that I have heard something ...


1

I would guess, and it can only be a guess, that Stewart is referring to weak measurement. There is a rather vague description of this in New scientist. Annoyingly I can't track down the original paper, but if Stewart's book was written in 2013 the timing fits.


1

The saucer ('cup') contains the ice at the beginning, which will melt so it contains water. As we are interested in what the scale shows us - the change in mass - it is not important whether it's ice or not, the mass would not change. The change we could observe is from evaporating water. How quick the water evaporates depends on the surface area, the ...


1

The mass of the melted water will be slightly lower than the ice. This is because the Latent Heat of Evaporation is higher than the Latent Heat of Fusion, so more water will be evaporated as it is turns into the liquid form and hence more mass will be lost. So gradually, as the water melts, the rate at which the mass is lost increases and hence the graph ...


1

Fast rotation keeps the paper-symmetry protons equivalent. High concentration makes for fast proton exchange in the alcohols (sharp line, no coupling to methylene protons). Integration identifies populations. The difference in chemical shifts tells you the temperature of the sample, ...


1

UPDATE I finally got it right. My thanks goes to @NeuroFuzzy, who pointed me in the right direction. According to wiki's Legendre polynomial solution for the elliptic integral, "an exact solution to the period of a pendulum is:" $$T=2\pi\sqrt\frac{\ell}{g}\sum\limits_{n=0}^\infty ...


1

This is really a comment to Mathbreaker's answer, but it's hard to do formulae in comments. If you simply solve: $$ T_m = 2\pi\sqrt{\frac{\ell}{g}}(1+\frac{1}{4}\sin^2(\tfrac{\alpha}{2})) $$ for $g$ you get: $$ g = \frac{4 \pi^2 l (4 + \sin^2(\tfrac{\alpha}{2}))^2}{16 \tau^2} \tag{1} $$ We use the identity: $$ \sin^2(\tfrac{\alpha}{2}) = \tfrac{1}{2} - ...


1

Engineering ToolBox has a rough empirical formula for predicting evaporation rates in kilograms per hour, $$g_h=\Theta A(x_s-x)$$ where $\Theta=25+19v$ is a dimensionless constant and where $v$ is the air velocity in m/s (but get rid of the units), where $A$ is the exposed water surface area in m$^2$, where $x_s$ is the humidity ratio of fully-saturated air ...



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