# Tag Info

## Hot answers tagged experimental-physics

15

Your worry is not necessary. In the usual experiment the detector measures the distribution of the time between the muon stopping in the detector and the time of it's decay. Then an exponential curve is fit to the data and the lifetime taken from the fit parameters Muon decay is a processes analogous to radioactive decay, and (like all exponential ...

6

The real issue is that the cup wasn't really full so that adding anything more would make it spill. You can clearly see the the level slowly growing above the top of the cup, as would be expected due to surface tension. Eventually another coin finally exceeded the limit, and a little water spilled. There is really nothing extraordinary going on here. ...

5

Given your description, you clearly have non-exponential behaviour. However, there are two possible reasons for this behaviour: Some materials in the system are nonlinear and do not follow Newton's law of cooling, which is that the heat flux at a given point is proportional to the temperature gradient vector. I should think this is the least likely of the ...

4

The last digits in 1000 are absolutely significant, they state that you have not 1200, not even 1001 but exactly 1000. In scientific notation, you would write this as $1.000 \times 10^3$. Compare this to $1\times10^3$ where you have just one significant digit. Update: consider the example from the question $1\times10^3+1.0$. The first term could be anything ...

3

To address the actual question of how we know the composition of UHECR without relying on source information (of which we have none), we have to look at their extensive air showers (EAS). After an UHECR hits the top of the atmosphere an EAS is created in the air, but p and Fe will create EAS with different shapes. Properties of hadronic interactions are ...

3

As long as your ambient light is consistent and lower power than your experiment light source it shouldn't be a problem. As you rotate the angle of the polarisation filter you will eventually detect a peak at the detector--this angle will be different from the polarisation angle when the solenoid is turned off, indicating Faraday rotation. Like most optics ...

3

Measuring dead-time (or other hardware efficiencies) is a non-trivial proposition, and there is no completely general solution. The answer that John gives in the comments ($\tau \times \text{number of events}$) is the best case: a system with few interconnections and no "extensible" contributions to the dead time. "Extensible" describes a system where ...

3

The curvature of the Earth is not enough to really 'curve' any object that's city-sized or smaller. The reason for this is that the radius of the Earth is really big.[citation needed] To make this more precise, say you make your chair one metre wide, and you have the edges pointing towards the centre of the Earth. You can then form an isosceles triangle ...

2

"The difference between theory and practice is that, in theory, there's no difference, but in practice, there is." At Colorado State University over the past several years, students using the Teachspin muon lifetime apparatus have been consistently getting muon lifetimes that are slightly but significantly too short. In late 2012 I spent several months ...

2

People defined time as a variable going forward long before entropy was defined. Biological/consciousness time, which you also discuss, forced the concept of time and a way /unit to measure it, as cultivation and buildings forced a concept of space and units to measure it. The celestial clock of sun moon and planets was used even by primitive people. ...

2

Couder and Fort's experiment is based on a mathematical analogy between the Hilbert space of a particle moving in two dimensions, and the two surface of a vibrating oil bath, which interacts with an oil droplet bouncing on top of it. Naïvely, one might try to extend this analogy a two-particle system by having two oil droplets bouncing on a single ...

2

You are right in that gravity did not change during data collection. You are a victim of uncertainty, which is a very important part of experimental physics. I'm sorry in advance for the "wall of text", and I hope that this clears up some confusion. The problem is that $1.50$ may not be exactly $1.500000000...$. Because the numbers are provided rounded, ...

1

I agree with kevinsa5 that the variation in $a$ is due to rounding errors, but I'd like to suggest a better way to analyse the data. Generally speaking, the best way to analyse data is to find a way to convert it to a straight line, then you can graph it and do a linear regression. In this case the way to procede is to note that if the acceleration is ...

1

I think you don't need to worry about the effect of the ambient light assuming the source and detector you used are the same (or roughly the same) as in the specifications, and you used the non-standard strategy (i.e. setting the relative angle between the polarizers to be $\pi/4$) as described in the "Experiments" section the ambient light was not ...

1

However, what if the time series is multi-dimensional, indeed of the same dimension as the phase space, to begin with? Well, how would you know that your time series is of the same dimension as the phase space? Usually, because you already know the dynamical equations for your system (as for your pendulum). If you observe a real-life complex system, ...

1

I can't tell what your equation means, but from your words "which part of our planet matters more in gravity, the core, or what's around it", you can define $$X=\frac{M_c}{M}=0.29$$ where $M_c$ is the mass of the core and $M$ is the mass of Earth, as obtained from this link. Since gravity from a spherical body with varying radial density is only dependent ...

1

Working with significant digits is very prone to error, because it can be misleading. It is much better to work with explicit errors. So to rewrite your example with explicit errors: $\left(1.0 \pm 0.5\right) \times 10^3 + 1.00\pm0.05$ We now add the errors quadratically (assuming they are uncorrelated): \$\sqrt{(0.5\times ...

1

Firstly, if your waveguide is a hollow conductor, it cannot support TEM modes. There must be at least two separate (electrically insulated from one another) conductors in the waveguide's cross section for TEM modes to propagate. The reason is that the transverse field dependence of a TEM mode is the same as that of a static field, as I explain in detail in ...

1

He used what is called a torsion balance. His experimental method is outlined very nicely in this video. After Coulomb published the result of his work, however, it was debated as to whether his experiment really did provide enough evidence to support his claim that the force between two point charges really did follow the equation we now call Coulomb's ...

1

What do you mean by "the waves radiate like a cone"? Anyhow, you will have reflections from the the metal pieces all around and that will be received to some extent and interfere with your measurements. if you don't want that happen then cover the metal with absorbing material. Check out how an "anechoic chamber" is constructed. Note that a dipole be it ...

1

Finally the bottom line of question is, can we experimentally prove that energy, total energy is actually conserved ? (a Yes answer requires a detailed experiment with complete conservation and no loopholes) Elementary particle physicists have been doing this for more than sixty years. Conservation of energy is one of the main constraints that built ...

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