# Dz Sentence Examples

dz
• If we suppose that the force impressed upon the element of mass D dx dy dz is DZ dx dy dz, being everywhere parallel to the axis of Z, the only change required in our equations (I), (2) is the addition of the term Z to the second member of the third equation (2).

• These equations can be made to represent the state of convective equilibrium of the atmosphere, depending on the gas-equation p = pk =RA (6) where 0 denotes the absolute temperature; and then d9 d p R dz - dz (p) n+ 1' so that the temperature-gradient deldz is constant, as in convective equilibrium in (I I).

• The time rate of increase of momentum of the fluid inside S is )dxdydz; (5) and (5) is the sum of (I), (2), (3), (4), so that /if (dpu+dpu2+dpuv +dpuw_ +d p j d xdyd z = o, (b)` dt dx dy dz dx / leading to the differential equation of motion dpu dpu 2 dpuv dpuv _ X_ (7) dt + dx + dy + dz with two similar equations.

• These equations may be simplified slightly, using the equation of continuity (5) § for dpu dpu 2 dpuv dpuw dt dx + dy + dz =p Cat +uax+vay+waz?

• Let us apply the above theorem to the case of a small parallelepipedon or rectangular prism having sides dx, dy, dz respectively, its centre having co-ordinates (x, y, z).

• Let this rectangular prism be supposed to be wholly filled up with electricity of density p; then the total quantity in it is p dx dy dz.

• Hence the total flux is - (+ d2V d 2 V d2V dye + dz2) dy dz, dx2 and by the previous theorem this must be equal to 4'rrp dxdydz.

• Consider the integral W dx dy dz .

• I by the whole area B"DZ'VO under the isothermal 9"D and the adiabatic DZ', bounded by the axes of pressure and volume.

• Here, then, is a case specially adapted to the isotropy of the quaternion system; and Hamilton easily saw that the expression i d x +j - + k dz could be, like ix+jy+ kz, effectively expressed by a single letter.

• If we take the axis of z normal to either surface of the film, the radius of curvature of which we suppose to be very great compared with its thickness c, and if p is the density, and x the energy of unit of mass at depth z, then o- = f o dz, (16) and e = f a xpdz,.

• The surface-density of this stratum is a = cp. The energy per unit of area is e = f xpdz=cp(X' -4lrpe(o))+27rp'f c 0(z) dz+27rp fee(c - z)dz.

• Hence the surface-tension =e - =47rp 2 (f 0(z)dz - ce(c)).

• Integrating the first term within brackets by parts, it becomes - fo de Remembering that 0(o) is a finite quantity, and that Viz = - (z), we find T = 4 7rp f a, /.(z)dz (27) When c is greater than e this is equivalent to 2H in the equation of Laplace.

• If the density be a, the attraction between the whole of one side and a layer upon the other distant z from the plane and of thickness dz is 27r6 2 P(z)dz, reckoned per unit of area.

• The expression for the intrinsic pressure is thus simply K= 2 iro 2 f 1,G(z)dz (28) In Laplace's investigation o- is supposed to be unity.

• We may call the value which (28) then assumes Ko, so that as above Ko =27rf (z)dz.

• If a i, a 2 represent the densities of the two infinite solids, their mutual attraction at distance z is per unit of area 21ra l a fZ '(z)dz, (30) or 27ra l 02 0(z), if we write f 4,(z)dz=0(z) (31) The work required to produce the separation in question is thus 2 7ru l a o 0 (z)dz; (32) and for the tension of a liquid of density a we have T = a f o 0 (z)dz.

• In all cases to which it is necessary to have regard the integrated terms vanish at both limits, and we may write f o (z)dz = 3f2' z 3 4(z)dz, f o z(z)dz = 'a' z4 cb(z)d z; (36) so that Ko = 3 f o z3 ?(z) dz, To = \$?

• Remember to do the three checks; Airspace, Alti and DZ, before starting any exercises and to respect all other canopy flyers.