# Pq Sentence Examples

The number of partitions of a biweight

**pq**into exactly i biparts is given (after Euler) by the coefficient of a, z xPy Q in the expansion of the generating function 1 - ax.From the law of angular motion of the latter its radius vector will run ahead of

**PQ**near A,**PQ**will overtake and pass it at apocentre, and the two will again coincide at pericentre when the revolution is completed.Recalling the formulae above which connect s P4 and a m, we see that dP4 and Dp q are in co-relation with these quantities respectively, and may be said to be operations which correspond to the partitions (

**pq**), (10 P 01 4) respectively.Since dp4+(-)P+T1(p +q qi 1)!dd4, the solutions of the partial differential equation d P4 =o are the single bipart forms, omitting s P4, and we have seen that the solutions of p4 = o are those monomial functions in which the part

**pq**is absent.B is the point at which the effect is required, distant a+b from 0, so that AB= b, AP=s,

**PQ**ds.Generally if S denotes any closed surface, fixed in the fluid, M the mass of the fluid inside it at any time t, and 0 the angle which the outward-drawn normal makes with the velocity q at that point, dM/dt = rate of increase of fluid inside the surface, (I) =flux across the surface into the interior _ - f f

**pq**cos OdS, the integral equation of continuity.Let

**PQ**be the ordinate of the point P FIG.Then the locus of the intersection of

**PQ**and OM is the quadratrix of Dinostratus.Let ON(= OP cos 0) be the perpendicular on

**PQ**.Since the conditions in the region

**PQ**remain always the same, the momentum perpendicular to AB entering the region at Q is equal to the momentum perpendicular to AB leaving the region at P. But, since the motion at Q is along AB, there is no momentum there perpendicular to AB.AdvertisementAgain if

**PQ**be any segment of the beam which is free from load, Q lying to the right of P, we find FPFQ, MPMQ=--F.**PQ**; (12)If

**PQ**be a short segment containing an isolated load W, we have FeFi.=W, MQ=MP; (3) hence F is discontinuous at a p concentrated load, diminishing by an amount equal to the load as we pass the loaded point to the right, whilst M is continuous.The plane in question is called the null-plane of P. If the null-plane of P pass through Q, the null-plane of will pass through P, since

**PQ**is a null-line.It is often convenient to resolve the forces on an element

**PQ**In the case of a chain hanging freely under gravity it is usually convenient to formulate the conditions of equilibrium of a finite portion

**PQ**.AdvertisementFor consider any two particles at P and Q, acting on one another with equal and opposite forces in the line

**PQ**.For in time t the mutual action between two particles at P and Q produces equal and opposite momenta in the line

**PQ**, and these will have equal and opposite moments about the fixed axis.The path of a point P in or attached to the rolling cone is a spherical epitrochoid traced on the surface of a sphere of the radius OP. From P draw

**PQ**perpendicular to the instantaneous axis.Let

**PQ**bubble.Conceive a perpendicular

**PQ**to be dropped from this point on the fundamental plane, meeting the latter in the point Q;**PQ**will then be parallel to OZ.AdvertisementIf we bisect AB and CD in P and Q respectively, and describe the arc

**PQ**of a circle with centre 0, the length of this arc is 2(b+a)0; and b - a =AB.