# Tag Info

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 ...

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 ...

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

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

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

$10\%$ 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

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 ...

2

As you stated in your question, the effect of an external magnetic field on an atom depends on the magnetic dipole moment of this atom. Before the introduction of spin, the only contributor to the magnetic dipole moment was the orbital dipole magnetic moment: $$\vec{M}=\vec{M_L}=-\frac{e}{2m_e}\cdot\vec{L}$$ which does not explain the S-G experiment since ...

1

If you want a function that has a certain graph, it's easy. First you digitize the graph, or ask the person who made the graph to send you the raw data. Then you define an interpolating function from this data-set. That's it! Now you have your function. I don't know why you would want to do this, but it is very easy to do. Usually in physics, we do not ...

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 ...

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 ...

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 ...

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), ...

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 ...

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