# ABC Sentence Examples

- If four fluids, a, b, c, d, meet in a point 0, and if a tetrahedron AB CD is formed so that its edge AB represents the tension of the surface of contact of the liquids a and b, BC that of b and c, and so on; then if we place this tetrahedron so that the face
**ABC**is normal to the tangent at 0 to the line of concourse of the fluids**abc**, and turn it so that the edge AB is normal to the tangent plane at 0 to the surface of contact of the fluids a and b, then the other three faces of the tetrahedron will be normal to the tangents at 0 to the other three lines of concourse of the liquids, an the other five edges of the tetrahedron will be normal to the tangent planes at 0 to the other five surfaces of contact. - Hence in trilinear co-ordinates, with
**ABC**as fundamental triangle, its equation is Pa+Q/1+R7=o. - Let us imagine unit mass of solution of volume V confined in a cylinder
**ABC**between a fixed vapour sieve B and a solid piston A A B C FIG. - = constant, _ ff 00 NdA N BA-AA X - JA (a' +X) (b 2 +A)P -
**abc**' a2 -b2 ' and at the surface A = o, I I N Bo-A 0 N I R - (a2+b2)**abc**a 2 -b 2**abc**a2b2 I /b 2 N = R I /b2 - I /a2**abc**I 1 I Bo - AO' a 2 b 2 - a2 b2 a 2 b2 = R (a 2 - b 2) /(a 22 + /b2) 2 - r (B o - Ao) U Bo+Co - B I - CI' Since - Ux is the velocity function for the liquid W' filling the ellipsoid A = o, and moving bodily with it, the effective inertia of the liquid in the interspace is Ao+B1+C1 Bo+Co - B1 - C, If the ellipsoid is of revolution, with b=c, - 2 XBo - - C BI' and the Stokes' current function 4, can be written down (I) is (5) (7) (6) The velocity function of the liquid inside the ellipsoid A=o due to the same angular velocity will be = Rxy (a2 - b2)/(a2 + b2), (7) and on the surface outside _ N Bo -Ao c1)0xy**abc**2 62' so that the ratio of the exterior and interior value of at the surface is ?o= Bo-Ao (9) 4)1 (a 2 -6 2)/(a2 + b) - (Bo - Ao)' and this is the ratio of the effective angular inertia of the liquid, outside and inside the ellipsoid X = o. - It is easily shown that the areas of the lune Adbea and the triangle
**ABC**are equal. - 7,
**ABC**is any triangle 1 Eisenlohr, Ein math. - The Hessian is symbolically (
**abc**) 2 azbzcz = H 3, and for the canonical form (1 +2m 3)xyz-m 2 (x 3 +y 3 +z 3). - - Germinating pollengrain of Epilobium (highly mag.) bearing a pollen-tube s; e, exine; intine;
**abc**, the three spots where the exine is thicker in anticipation of the formation of the pollen-tube developed in this case at a. - = constant, _ ff 00 NdA N BA-AA X - JA (a' +X) (b 2 +A)P -
**abc**' a2 -b2 ' and at the surface A = o, I I N Bo-A 0 N I R - (a2+b2)**abc**a 2 -b 2**abc**a2b2 I /b 2 N = R I /b2 - I /a2**abc**I 1 I Bo - AO' a 2 b 2 - a2 b2 a 2 b2 = R (a 2 - b 2) /(a 22 + /b2) 2 - r (B o - Ao) U Bo+Co - B I - CI' Since - Ux is the velocity function for the liquid W' filling the ellipsoid A = o, and moving bodily with it, the effective inertia of the liquid in the interspace is Ao+B1+C1 Bo+Co - B1 - C, If the ellipsoid is of revolution, with b=c, - 2 XBo - - C BI' and the Stokes' current function 4, can be written down (I) is (5) (7) (6) The velocity function of the liquid inside the ellipsoid A=o due to the same angular velocity will be = Rxy (a2 - b2)/(a2 + b2), (7) and on the surface outside _ N Bo -Ao c1)0xy**abc**2 62' so that the ratio of the exterior and interior value of at the surface is ?o= Bo-Ao (9) 4)1 (a 2 -6 2)/(a2 + b) - (Bo - Ao)' and this is the ratio of the effective angular inertia of the liquid, outside and inside the ellipsoid X = o. - 6,
**ABC**is an isosceles triangle right D FIG. - Right angled at C, semicircles are described on the three sides, thus forming two lunes Afcda and Cgbec. The sum of the areas of these lunes equals the area of the triangle
**ABC**.] As for Euclid, it is sufficient to recall the facts that the original author of prop. 8 of book iv. - 2, in which the ordinates of the line
**ABC**represent the percentage of pearlite corresponding to each percentage of carbon, and the intercept ED, MN or KF, of any point H, P or L, FIG. - Hastening its cooling by casting it in a cool mould, favours the formation of cementite rather than of graphite in the freezing of the eutectic at
**aBc**, and also, in case of hyper-eutectic iron, in the passage through region 3. - Malik al-Adi~ Saif al-din
**AbC**Bakr, 596615 (1199-1218). - Thus if the three lines form a triangle
**ABC**, and if the given force F meet BC in H, then F can be resolved into two components acting in HA, BC, respectively. - It is generated as follows: Let
**ABC**be a circle having AB for a diameter. - If the right-angled triangle
**abc**, fig. - A, b, c a' b' c', a"b" c", b", that is, every term which does not vanish contains as a factor the
**abc-determinant**last written down; the sum of all other factors a0'y" is the a/37-determinant of the formula; and the final result then is, that the determinant on the left-hand side is equal to the product on the right-hand side of the formula. - In order to explain the Peltier effect, Kohlrausch further assumes that an electric current, C, carries a heatflow, Q=
**ABC**, with it, where " A is a constant which can be made equal to unity by a proper choice of units." - Another series of instruments, introduced by Cooke and Wheatstone in 1840, and generally known as " Wheatstone's step-by-step letter-showing " or "
**ABC**instruments," were worked out with great ingenuity of detail by Wheatstone in Great Britain and by Breguet and others in France. - We can see that (
**abc**)a x b x c x is not a covariant, because it vanishes identically, the interchange of a and b changing its sign instead of leaving it unchanged; but (**abc**) 2 is an invariant. - It is remarkable that Mobius employs the symbols AB,
**ABC**,**Abcd**In Their Ordinary Geometrical Sense As Lengths, Areas And Volumes, Except That He Distinguishes Their Sign; Thus Ab = Ba,**Abc**= Acb, And So On. - 3 the curve of brightness for one component is
**ABC**, and for the other OA'C'; and the curve representing half the combined brightnesses is E'BE. - Take any two arbitrary directions in the plane of the paper, and draw a small isosceles triangle
**abc**, whose sides are perpendicular to the two directions, and consider the equilibrium of a small triangular prism of fluid, of which the triangle is the cross section. - A scalene triangle
**abc**might also be employed, or a tetrahedron. - If at any points of a vortex line the cross-section
**ABC**, A'B'C' is drawn of the vortex filament, joined by the vortex line AA', then, since the flow in AA' is taken in opposite directions in the complete circuit**ABC**AA'B'C' A'A, the resultant flow in AA' cancels, and the circulation in**ABC**, A'B'C' is the same; this is expressed by saying that at all points of a vortex filament wa is constant where a is the cross-section of the filament and w the resultant spin (W. - The extension to the case where the liquid is bounded externally by a fixed ellipsoid X= X is made in a similar manner, by putting 4 = x y (x+ 11), (io) and the ratio of the effective angular inertia in (9) is changed to 2 (B0-A0) (B 1A1) +.a12 - a 2 +b 2 a b1c1 a -b -b12
**abc**(Bo-Ao)+(B1-A1) a 2 + b 2 a1 2 + b1 2 alblcl Make c= CO for confocal elliptic cylinders; and then _, 2 A? ? **ABC**is a quadrant in which the line AB and the arc AC are divided into the same number of equal parts.- If now we cut the freezing-point surface by planes parallel to the base
**ABC**we get curves giving us all the alloys whose freezing-point is the same; these isothermals can be projected on to the plane of the triangle and are seen as dotted lines in fig. - Also, let angle
**ABC**=7r - 0, angle BCD =ir - 4, angle between BC and AD = G. - If now the curve moves along unchanged in form in the direction
**ABC**with uniform velocity U, the epoch e =OA at any time t will be Ut, so that the value of y may be represented as 2 y=a sin T (x - Ut). - Aziz
**AbC**MansCr Nizar (al-Aziz billgh), 365386 (975-996). - By adjusting the right ascension of the plane
**ABC**and rotating the axis with the angular velocity of the sun, it follows that BC will be the direction of the solar rays throughout the day. - The centre of the circle
**ABC**, through an angle AJB. - About OA, OB, OC, respectively, is to leave the body in its original position, provided the circuit
**ABC**is left-handed as seen from 0. - The two forces at B will cancel, and we are left with a couple of moment P.AC in the plane AC. If we draw three vectors to represent these three couples, they will be perpendicular and proportional to the respective sides of the triangle
**ABC**; hence the third vector is the geometric sum of the other two. - Be situate at the vertices of a triangle
**ABC**, the mass-centre of ~ and y is at a point A in BC, such that ~. - Also, by giving suitable values (positive or negative) to the ratios a :7 we can make G assume any assigned position in the plane
**ABC**. We have here the origin of the barycentric co-ordinates of MObius, now usually known as areal co-ordinates. - For if in the triangle
**abc**the side ab is taken so as to represent on a given scale the tension of the surface of contact of the fluids a and b, and if the other sides be and ca are taken so as to represent on the same scale the tensions of the surfaces between b and c and between c and a respectively, then the condition of equilibrium at 0 for the corresponding tensions R, P and Q is that the angle ROP shall be the supplement of**abc**, POQ of bca, and, therefore, QOR of cab. - If four fluids, a, b, c, d, meet in a point 0, and if a tetrahedron AB CD is formed so that its edge AB represents the tension of the surface of contact of the liquids a and b, BC that of b and c, and so on; then if we place this tetrahedron so that the face
**ABC**is normal to the tangent at 0 to the line of concourse of the fluids**abc**, and turn it so that the edge AB is normal to the tangent plane at 0 to the surface of contact of the fluids a and b, then the other three faces of the tetrahedron will be normal to the tangents at 0 to the other three lines of concourse of the liquids, an the other five edges of the tetrahedron will be normal to the tangent planes at 0 to the other five surfaces of contact.