New answers tagged

2

I think your problem is that you didn't change the units in the constant g. It has a value of approximately $9.8ms^{-2}$. Notice that it depends on meters. To obtain the correct result, you should use $980cms^{-2}$. Notice that this constant is off by a factor of 100, so that the result (after the square root) is off by a factor of $\sqrt{100}=10$.


0

If you're setting $\hbar=1$, then you don't - you can't - distinguish between energy and angular frequency. They are, in fact, the same quantity, since $E=\hbar\omega=\omega$. Similarly, if $\hbar=1$ you can no longer draw dimensional distinctions between wavevectors and linear momenta, or between angular momenta and pure numbers. In general, you only set ...


-1

In the case of energy, I find it more useful to use the reciprocal of time rather than frequency. Time is a much more fundamental (and common) dimension. In other words $E=1/T$, and of course, $T=1/E$.


0

(Thanks to Drebin J.) Let's suppose there's a building with many floors, labelled G, 1-10, and after that continues with A1, A2... A10, B1,B2 etc. A1 is floor 1 but actually the 11th, while A2 is 2nd but actually 12th. (Similar to Celsius in Kelvin) Supposing I jump from A2 to A1 (from the staircase), it would obviously hurt. The damage incurred ...


1

In an exam, Alice scored $50$, Bob scored $40$, Eve scored $10$. Now to raise the class mean, the teacher decided to add $20$ points to every students. So Alice's score becomes $70$, Bob's score becomes $60$, and Eve's score becomes $30$. Now Alice complained to the teacher, "my score was higher than Bob's by $10$ marks, and you see, according to the ...


0

This seems to me like the general "Problem", that differences of Observables behave in a different way, than the observables itself do: One other example (that has nothing to do with additivity or multiplicativity, as you stated) are differences of frequencies: You know that you can convert a frequency to a wavelength by dividing c by the frequency: ...


1

Celsius and Kelvin are two scales that differs only for an additive factor, but the single increment corresponds to the same temperature difference. In other words, an object become "hotter" in the same way if you rise its temperature by 1K or 1°C. You can use conversion formula in differences, just make sure you use it for both terms and keep in mind that ...


0

A temperature change of 1$^\circ$C is the same as a temperature change of 1K. So if you start at 30$^\circ$C (= 303.15K ) and increase the temperature by 5$^\circ$C (5K) the new temperature is 30 + 5 = 35$^\circ$C (303.15 + 5 = 308.15K).


0

temperature difference is 5°C, which is convertible to 278.15K. This is where you go wrong. A difference of 5°C is a difference. Express it as you will in Celcius, Kelvin, Rankine or even in Fahrenheit. But it is a difference. Saying 5°C, which is convertible to 278.15K mean that you are not looking at a difference. Instead you are calculating how to ...


1

Of course, you can do it with Matlab, Mupad, Maple, Mathematica or even the Smart Math Calculator. Use this method: First define your variables with your units of choice, then tell the programm what the conversion factors from the given units to the target units are, for example, if you have km/h and need m/sec define 1km as 1000m and 1sec as h as 60² ...


1

There are two mistakes. As AccidentalFourierTransform pointed out, the coefficient $7.181\times 10^{-16}$, when converted from MeV to eV, should give $7.181\times 10^{-46}$. Mega means a million, and it to the fifth power gives $10^{30}$, not just $10^{15}$. In this way, the OP has to add a $10^{-15}$ factor to his result. That makes his result $10^{-3}$ ...



Top 50 recent answers are included