# Dt sentence example

dt
• Then the first of the equations of motion may be put under the form dt ?
• According to this notation, the three equations of motion are dt2 = b2v2E + (a2 - b2) d.s dt =b2v2rj+(a2 - b2) dy d2 CIF - b2p2+(a2_b2)dz It is to be observed that denotes the dilatation of volume of the element situated at (x, y, z).
• 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?
• So far these theorems on vortex motion are kinematical; but introducing the equations of motion of § 22, Du + dQ =o, Dv+dQ =o, Dw + dQ dt dx dt dy dt dz and taking dx, dy, dz in the direction of u, v, w, and dx: dy: dz=u: v: w, (udx + vdy + wdz) = Du dx +u 1+..
• Thus if T is expressed as a quadratic function of U, V, W, P, Q, R, the components of momentum corresponding are dT dT dT (I) = dU + x2=dV, x3 =dW, dT dT dT Yi dp' dQ' y3=dR; but when it is expressed as a quadratic function of xi, 'x2, x3, yi, Y2, Y3, U = d, V= dx, ' w= ax dT Q_ dT dT dy 1 dy2 dy The second system of expression was chosen by Clebsch and adopted by Halphen in his Fonctions elliptiques; and thence the dynamical equations follow X = dt x2 dy +x3 d Y = ..., Z ..., (3) = dt1 -y2?y - '2dx3+x3 ' M =..
• He therefore employed the corresponding expression for a cycle of infinitesimal range dt at the temperature t in which the work dW obtainable from a quantity of heat H would be represented by the equation dW =HF'(t)dt, where F'(t) is the derived function of F(t), or dF(t)/dt, and represents the work obtainable per unit of heat per degree fall of temperature at a temperature t.
• After a time dt the value of p i will have increased to pi+pidt, where p i is given by equations (i), and there will be similar changes in qi, P2, q2, ...
• Thus after a time dt the values of the coordinates and momenta of the small group of systems under consideration will lie within a range such that pi is between pi +pidt and pi +dp,+(pi+ap?dpi) dt „ qi +gidt „ qi+dqi+ (qi +agLdgi) dt, Thus the extension of the range after the interval dt is dp i (i +aidt) dq i (I +?gidt).
• Each of the molecules enumerated in expression (9) will move parallel to the edge of this cylinder, and each will describe a length equal to its edge in time dt.
• Thus each of these molecules which is initially inside the cylinder, will impinge on the area dS within an interval dt.
• The cylinder is of volume u dt dS, so that the product of this and expression (9) must give the number of impacts between the area dS and molecules of the kind under consideration within the interval dt.
• Thus the contribution to the total impulsive pressure exerted on the area dS in time dt from this cause is mu X udtdS X (11 3 m 3 /,r 3)e hm (u2+v2+w2 )dudvdw (I o) The total pressure exerted in bringing the centres of gravity of all the colliding molecules to rest normally to the boundary is obtained by first integrating this expression with respect to u, v, w, the limits being all values for which collisions are possible (namely from - co too for u, and from - oo to + oo for v and w), and then summing for all kinds of molecules in the gas.
• The aggregate amount of these pressures is clearly the sum of the momenta, normal to the boundary, of all molecules which have left dS within a time dt, and this will be given by expression (pp), integrated with respect to u from o to and with respect to v and w from - oo to +oo, and then summed for all kinds of molecules in the gas.
• Now if the notch of the tan gent sight be carried to H' in order to lay on T, the fore-sight, and with it the axis, H will be moved to F', the line of fire will be HF'D', and the shot will strike T since D'T = DT.
• In the time dt which the wave takes to travel over MN the particle displacement at N changes by QR, and QR= - udt, so that QR/MN = - u/U.
• Generally, if any condition in the wave is carried forward unchanged with velocity U, the change of 4 at a given point in time dt is equal to the change of as we go back along the curve a distance dx = Udt at the beginning of dt.
• If P is the undisturbed pressure and P+w the pressure at AB, the momentum entering through AB per second isJ01(P+w-+pu2)dt.
• The excess pressure on CD is therefore 4 1 (c:3+ pu 2)dt.
• Rods of different materials may be used as sounders in a Kundt's dust tube, and their Young's moduli may be compared, since: length of rod Then dO U = - ax dx or dt = - UK.
• Let us Freezing freeze out unit mass of solvent from a solution at its freezing point T - dT and remove the ice, which is assumed to be the ice of the pure solvent.
• Then let us heat both ice and solution through the infinitesimal temperature range dT to the freezing point T of the solvent, melt the ice by the application of an amount of heat L, which measures its latent heat of fusion, and allow the solvent so formed to enter the solution reversibly through a semi-permeable wall into an engine cylinder, doing an amount of work Pdv.
• By cooling the resultant solution through the range dT we recover the original state of the system.
• But we have seen that the depression of dT of the freezing point of a dilute solution is measured by TPdv/L.
• The latent heat L at any temperature is given by L=Lo - f 0 64 0 (s - s')dT, where Lo is value at To and s--s' is the difference in the specific heats of water and ice.
• Now that friction is also the difference between the tensions of the band at the two ends of the elementary arc, or dT =dF =fTdO; which equation, being integrated throughout the entire arc of contact, gives the following formulae:
• Let da be the deviation of angular velocity to be produced in the interval dt, and I the moment of the inertia of the body about an axis through its centre of gravity; then 1/8Id(&) = Iada is the variation of the bodys actual energy.
• Let M be the moment of the unbalanced couple required to produce the deviation; ther by equation 57, 104, the energy exerted by this couple in tht interval dt is Macit, which, being equated to the variation of energy gives da R2W da -
• We have therefore dT = 1.1Td9 and dT/T =µd9.
• The differential equations which completely determine the changes in the co-ordinates x and y, or the motion of m relative to M, are: - d2x (M+m)x di' r3 (1) d2y _ (M+m)y dt 2 r3 These formulae are worthy of special attention.
• Putting a, b, c, d, for the constants, the general form of the solution will be x = fl (a,b,c,d,t) y = f2(a,b,c,d,t) From these may be derived by differentiation as to t the velocities dt =f '1(a,b,c,d,t) = x'  ?
• But in every case, if at a certain time t, the actual planet has a certain longitude, it is certain that at a very short interval dt before or after t, the fictitious planet will have this same longitude.
• What Hansen's method does is to determine a correction dt such that, being applied to the actual time t, the longitude of the fictitious planet computed for the time t+dt, will give the longitude of the true planet at the time t.
• By a number of ingenious devices Hansen developed methods by which dt could be determined.
• If the quantity of heat absorbed and converted into electrical energy, when unit quantity of electricity (one ampere-second) flows from cold to hot through a difference of temperature, dt, be represented by sdt, the coefficient s is called the specific heat of electricity in the metal, or simply the coefficient of the Thomson effect.
• The element might thus be regarded as the seat of an E.M.F., dE=sdT, where dT is the difference of temperature between its ends.
• On this hypothesis, if we confine our attention to one of the two metals, say p", in which the current is supposed to flow from hot to cold, we observe that p"dT expresses the quantity of heat converted into electrical energy per unit of electricity by an E.M.F.
• By applying the first law of thermodynamics, Kohlrausch deduces that a quantity of heat, CBdT, is absorbed in the element dT per second by the current C. He wrongly identifies this with the Thomson effect, by omitting to allow for the heat carried.
• I believe my input so far has been useful to both the executive directors and the DT Board.
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