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The reason why the tuning fork creates a sound is because it makes the air molecules around it vibrate, which means that there is a longitudinal sound wave created. If I understand correctly, the bottom of your tuning fork was placed into a wooden block and then struck with a hammer. The wooden block does not actually resonate (the frequency of the tuning ...


0

See What Really Happens in the Franck-Hertz Experiment with Mercury? The article explains how the curve depends upon the design of the tube.


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Just a guess, but perhaps he actually said (or meant) "1 cubic centimeter". Increasing the rod length by 10% is not the same as adding 1 cubic centimeter (because the rod is circular, not square, in the other dimensions). Also assuming the diameter of the rod is 1cm.


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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 $11 \text{ cm} \pm 0.11 \text{ cm}$. ...


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$10\%$ 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 ...


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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 ...


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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, ...


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As noted within my other post at Time dilation in special relativity, here I have basically the same situation yet the box becomes a 300,000 km long spaceship. The basics are explained. However, the only absolute measure that exists within it, is that all objects are constantly traveling at the speed of light within the 4 dimensional environment known as ...


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Notice that the movement of light must always be treated as a local phenomenon, which in your thought experiment means the light is travelling only within the moving box. It is not travelling from the moving box to a stationary observer. Therefore, throughout the whole experiment the box is stationary for the light and no wall is moving toward or from it. ...


1

Doppler shift occurs only when the sender, the receiver or both are moving relatively to each other. As the black boxes rest at the bottom of the ocean and the search ships move relatively slow, there won't be any significant Doppler shift. However, if the Ocean Shield receives several signals at different locations (the location of the Ocean Shield), the ...


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Most campuses have a ceramics section to the art department. Gas and electric kilns are great for this. At the temperatures called "cone 9" and "cone 10" everything inside the kiln becomes invisible. It is just red-orange inside. The light bulb is a terrible example. The filament in an insulating chamber with a tiny hole is OK as long as the whole chamber ...


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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.


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The key point that you are missing is that the speed of light is constant for all inertial frames of reference. If you are going $0.99\ c$ and you are holding a flashlight and you turn it on, the photons emitted from the flashlight will appear to you to be leaving the flashlight at exactly $c$ (the speed of light). The key point of special relativity is ...


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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. ...


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In a sphere, any light emitted from the center will reflect off the sides at normal incidence come back to the center. In a cube, some rays never return to the center, so you aren't measuring all of the light emitted, which defeats the purpose of the device.


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If a cold beer can is taken out of the fridge on a warm day small water droplets form on its surface since the surface temperature is below the dew point of the moist ambient air. After a while, when the can reaches equilibrium temperature the water gets evaporated away again. For the ice cube the same is true in principle besides it evaporates away ...


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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 ...


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The matter is constant. The mass increases (latent enthalpy of fusion plus specific heat as temperature rises) but ${m = E/c^2}$, hence way too small to measure. What is big enough to measure is the deceasing net buoyancy for displaced air by the denser mass (lesser volume) of of water versus ice, about 1.3 ${mg/cm^3}$ differential air volume at STP. If ...


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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 ...


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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 ...


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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 ...


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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 $70\text{ kg}$, ...


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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 ...


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when you go fast enough the water molecules just can't move out of the way fast enough for a soft landing.


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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 ...


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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.


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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 ...


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One way would be to use a radioactive tracer gas. Krypton-85 has been used to study atmospheric flow: http://onlinelibrary.wiley.com/doi/10.1029/JC075i015p02985/abstract Otherwise, spectroscopic techniques such as absorbance or emission spectra could be used to measure the concentration of various gases at different points in space and time.


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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} - ...


0

$$g= \frac{L *4\pi^2 * ((5- \cos^2 (\alpha/2))^2}{16T^2}$$ This is what i got when i simplified for $g$. I assume you made some mistake in simplification, or it just might be me, Iapologize if it is. You could try plugging your values into this and hope for an answer... when i simplify the latter part as well i get $$\frac{L * 4\pi^2 ...


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:


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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 ...


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 ...


1

This is an excellent question. At the heart of it is this: you compare the clock to another copy of the same clock. Well, actually you need to compare three identical clocks to each other to make a strong statement about the clock noise, but lets not worry about that for now. If the noise of your clock is stationary (which it better be for a good clock), ...


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Let's first recall that there is a strict definition for (how to determine) whether a given clock $A$ is "accurate" (or "good"); namely: if for any three of its indications, $A_J$, $A_K$ and $A_Q$, the durations of clock $A$ between pairs of those indications, say $\Delta \tau_A [ \small{\text{from }} A_J \small{\text{ until }} A_K ]$ and $\Delta \tau_A [ ...


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, ...


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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 ...


1

Since Community in its wisdom brought this up again, lets make things clear: 1.Suppose if we have the planning of home to be made in the same manner of the circle shaped fibre with same material as that of quartz. We can have same light circulating all over once incident. So, if this happens we can have light through out the house with out light being ...


1

@AwalCraig, In the case of a loosened cable string being struck, as opposed to being cut (as @Craig pointed out) is the effect of the momentum of the object striking the cable/string being lost as the loose object dissipates this energy. You see this in car races along the edges that use hay bails, or barrels of water, and behind those highway trucks that ...


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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 ...


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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.


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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 ...


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: ...


1

The answer is Yes. However the losses will be wavelength-dependent, you will likely get a "ugly" superposition of modes at the output and a dispersed signal as user288447 stated. You are correct, the mode for each wavelength need to be solution of the Helmholtz equation. For a dielectric waveguide, there is no absolute cut-off wavelength (all wavelength can ...


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Obviously, the smallest particle that scientists have ever seen directly is a photon. The question is a bit silly because it tries to eliminate simple device like a photographic plate, but the human eye, its nerves and the visual cortex together are far more complicated.


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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 ...


-1

Measurement of length or dimension relates to human experience for quantifying and comparing . If human awareness is absent a point can be defined as infinite size or dimensionless or single or multidimensional.


0

I think you would underestimate the error if you used that calculation. For example, if you had a thermometer that the manufacturer said was accurate within 1 degree, and you took a reading every second for 100 years to calculate the average temperature in Chicago, the uncertainty in the average could still be about 1 degree, because the thermometer could ...


0

That's a fundamental problem with all measurement devices. What you can do is to compare your clock with other clocks. If yours differs significantly from the average then you know your clock is inaccurate. Of course, this does not protect you from overall errors (i.e. a fundamental flaw in the mechanics/physics behind the device).


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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 ...



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