Sentences with phrase «for kt»

does this work for kt 5.0.3?

He paid for the kt boundary dating construction of the Central American Justice.

Not exact matches

The recording of this material was enabled by a generous support of Heidelberg Institute for Theoretical Studies (HITS) and Klaus Tschira Stiftung (KTS), Heidelberg, Germany.

Project: Using methods of researching for exploring the Sun and other objects from the Solar System Authors: Tsvetoslav Nikolaev Nikolov; Ivan Nikolaev Velchev, Adrian Ivanov Grigorov Teacher: Tsetsa Tsolova Hristova School: PG po KTS, Pravets BULGARIA

We need relatively calm seas (e.g, < 10 - 15 kt winds and < 3 - 4 ft seas) and if you check out the offshore actual weather and forecasts for the outer waters where we hope to work with beaked whales and other pelagic -LSB-...]

Why you'll love it: Situated in the pastoral area of Canggu, KTS Day Spa and Retreat offers a great getaway with an outdoor swimming pool, free WiFi access throughout the property, and free on - site private parking for guests who drive.

Finds the maximum reduction in intensification time is noticed over the North Atlantic Ocean where the average time needed for TC to intensify from 64 kt to 112 kt has reduced by nearly 20 hours during the past 25 - year period

For example, in a solid there are only three axes, but there is KE and PE associated with each axis, so there are SIX degrees of freedom and each will have an average energy 1/2 kT when averaged over sufficiently long periods of time.

8 by N and separate out the terms in the brackets, we can see that the total kinetic energy basically comprises the ordinary thermal energy of the gas (f / 2)(N - 1) kT, independent of height, plus a term -LRB-- mgz) for the N = 1 projectile, close enough.

The Maxwell - Boltzmann distribution gives a relative occupation for any state with energy E at temperature T of simply exp -LRB-- E / kT).

If DFcr < > kT, and there's essentially no mathematical difference between the B - E and M - B statistics for the problem.

Start with the B - E probability for two different states E1 and E2, both much larger than kT.

Take the ratio of these probabilities and you get a formula extremely close to Boltzmann and not at all close to B - E for the term delta E / kT = (E2 - E1) / kT.

Plug in the B - E formula for state 1 (1 / (exp -LRB-(E1 — mu) / kT)-- 1) and state 2 (1 / (exp (E2 — mu) / kT)-- 1) and you'll see the key question is not the energy difference E2 — E1 vs kT, but what E1 — mu is relative to kT.

If there are a large number of quantum states available for a single particle, then that occupation number must be small, and therefore E1 — mu must be much larger than kT.

If E — mu >> kT even for the lowest energy state E, then you can ignore the «-1», the constant «mu» factors out as a constant multiplier and you reduce to a Maxwell - Boltzmann form.

Reduced the «turbulent conductivity» values as the surface was reached — instead of one «turbulent conductivity» value (used when the layer below was warmer than the layer above), these values were reduced closer to the surface, e.g. for the 100μm layer, kt = 10; for the 300μm layer, kt = 10; for the 1 mm layer, kt = 100; for the 5 mm layer, kt = 1000; for the 20 mm layer, kt = 100,000.

K — a very low value as water is a poor conductor of heat... a value for stirred water that I found in a textbook: kt = 2 x 105 W / m.

I used a value for stirred water that I found in a textbook: kt = 2 x 105 W / m.

Disagreement is healthy, but with our technology and that of Climeworks now being piloted at scales near 1 kt - CO2 / year, there is sufficient engineering for third parties do a far more realistic assessment than the APS was able to do.

/ c (j); % heat cap T (j, i) = T (j,i - 1) + dT; % new temperature end if T (1, i) > 1000 % FE problem disp -LRB-[«Terminated at period «num2str (p) «and i = «num2str (i)-RSB--RRB-; return end % now check for convective overturning and use the results in the next % iteration for j = 2: nlayers if T (j, i) > T (j - 1, i) kk (j) = kt; % turbulent or stirred conductivity else kk (j) = ks; % still conductivity end end

It is actually a shorthand for describing Maxwell - Boltzmann statistics applied to material properties: exp -LRB-- E / kT) where E is an activation energy.

In terms of the usual definition of KE (for which the average KE of a molecule = (3/2) kT), the straightforward result of applying the Virial Theorem to a homogeneous gravitational field is: (2/3)[KE] = [PE], as I have demonstrated previously (and also I have updated the text to recover a Word version, in the last few days).

For example, the Ku - band QuikSCAT radar can not measure extreme winds or winds in heavy rain (although it can measure wind speeds of up to about 90 kt, if those winds occur outside of rain and are not confined to a very small area, both of which are the case in most hurricanes).

Your statement: «The correct formula to calculate the intensity of the solar radiation is: I = -LSB-(2hv ^ 3 / c ^ 2)(1 / e ^ (hv / kT)-1)-RSB- * ε sun < / i» (which is incorrect, as pointed out before that is the Planck function which is the intensity for each wavelength)