## Sec Sentence Examples

- The declaration is to the effect that the clergyman has not received the presentation in consideration of any sum of money, reward, gift, profit or benefit directly or indirectly given or promised by him or any one for him to any one; that he has not made any promise of resignation other than that allowed by the Clerical Resignation Bonds Act 1828; that he has not for any money or benefit procured the avoidance of the benefice; and that he has not been party to any agreement invalidated by
**sec**. 3 sub-**sec**. 3 of the act which invalidates any agreement for the exercise of a right of patronage in favour or on the nomination of any particular person, and any agreement on the transfer of a right of patronage (a) for the retransfer of the right, or (b) for postponing payment of any part of the consideration for the transfer until a vacancy or for more than three months, or (c) for payment of interest until a vacancy or for more than three months, or (d) for any payment in respect of the date at which a vacancy occurs, or (e) for the resignation of a benefice in favour of any person. - Persons who have held the licence to trade (
**sec**FINANCE) for five years and upwards. - Taking two planes x = =b, and considering the increase of momentum in the liquid between them, due to the entry and exit of liquid momentum, the increase across dy in the direction Oy, due to elements at P and P' at opposite ends of the diameter PP', is pdy (U - Ua 2 r2 cos 20 +mr i sin 0) (Ua 2 r 2 sin 2 0+mr 1 cos 0) + pdy (- U+Ua 2 r 2 cos 2 0 +mr1 sin 0) (Ua 2 r 2 sin 2 0 -mr 1 cos 0) =2pdymUr '(cos 0 -a 2 r 2 cos 30), (8) and with b tan r =b
**sec**this is 2pmUdo(i -a 2 b2 cos 30 cos 0), (9) and integrating between the limits 0 = 27r, the resultant, as before, is 27rpmU. - Well as of the body from the vector OF to O'F' requires an impulse couple, tending to increase the angle F00', of magnitude, in
**sec**. foot-pounds F.00'.sin FOO'=FVt sin (0-0), (4) equivalent to an incessant couple N=FV sin (0-0) = (F sin 0 cos 0-F cos 0 sin ¢)V = (c 2 -c i) (V /g) sin 0 cos 4) =W'(13-a)uv/g (5) This N is the couple in foot-pounds changing the momentum of the medium, the momentum of the body alone remaining the same; the medium reacts on the body with the same couple N in the opposite direction, tending when c 2 -c 1 is positive to set the body broadside to the advance. - Consider, for example, a submarine boat under water; the inertia is different for axial and broadside motion, and may be represented by (1) c 1 =W+W'a, c2=W+W'/3' where a, R are numerical factors depending on the external shape; and if the C.G is moving with velocity V at an angle 4) with the axis, so that the axial and broadside component of velocity is u = V cos 0, v =V sin 4), the total momentum F of the medium, represented by the vector OF at an angle 0 with the axis, will have components, expressed in
**sec**. Ib, F cos 0 =c 1 - = (W +W'a) V cos 43, F sin 0 = c 2.11 = (W +W'/3) V sin 4) . - The moment of inertia of the body about the axis, denoted by But if is the moment of inertia of the body about a mean diameter, and w the angular velocity about it generated by an impluse couple M, and M' is the couple required to set the surrounding medium in motion, supposed of effective radius of gyration k', If the shot is spinning about its axis with angular velocity p, and is precessing steadily at a rate about a line parallel to the resultant momentum F at an angle 0, the velocity of the vector of angular momentum, as in the case of a top, is C i pµ sin 0- C2µ 2 sin 0 cos 0; (4) and equating this to the impressed couple (multiplied by g), that is, to gN = (c 1 -c 2)c2u 2 tan 0, (5) and dividing out sin 0, which equated to zero would imply perfect centring, we obtain C21 2 cos 0- (c 2 -c 1)c2u 2
**sec**0 =o. - The first, from the north, comprises the upper basins
**Sec**ti of the Maranon and the Huallaga, and is 350 m. - 3.019 cm./
**sec**. 3.009 cm./**sec**. 2.9815 cm./**sec**. 2.9993 CM. - /
**sec**. 2 '9955 cm./**sec**. 2.987 cm./**sec**. 3.009 CM. - /
**sec**. 2' 99 2 cm./**sec**. 3.002 cm. - The
**sec**and phase of the offensive as planned could not even be commenced. - This discourse, from its explanatory character, and from the easy conversational manner of its delivery, was for a long time called o 1 u Xia rather than Aoyos: it was regarded as part of 1 See Philo, Quod omnis probes liber,
**sec**. 12 (ed. - U 0 = 28,000 cm.
**sec**. (about 920 ft./**sec**.). - Newton found 979 ft./
**sec**. But, as we shall see, all the determinations give a value of Uo in the neighbourhood of 33, 000 cm./**sec**., or about 1080 ft./**sec**. This discrepancy e was not explained till 1816, when Laplace (Ann. - Uo=33,150 cm./
**sec**., which is closely in accordance with observation. - This gives maximum u=about 8 cm./
**sec**., which would not seriously change the form of the wave in a few wavelengths. - Mag., 1879, 7, p. 219) investigated the transmission of a report from a cannon in different directions; he found that it rose to a maximum of 1267 ft./
**sec**. at 70, to 90 ft. - 1280 metres arid 2445 metres, obtaining from the first U 0 =331.37 met./
**sec**.; but the number of experiments over the longer distance was greater, and he appears to have put more confidence in the result from them, viz. - Was Uo=330 6 met./
**sec**., while for a diameter 0.108 it was U 0 =324' 25 met./**sec**. - The velocity deduced at 8.1° C. was U=1435 met./
**sec**., agreeing very closely with the value calculated from the formula U 2 = E/p. - 5,
**sec**. i of the state constitution, authorized the initiative and referendum, but twofifths of the entire number of counties must each furnish for initiative petitions signatures amounting in number to 8% of the whole number of votes cast for governor at the election last preceding the filing of the petition; for referendum petitions two-fifths of the counties must each furnish as signers 5% of the legal voters; and any measure referred to the people shall be in full force unless the petition for the referendum be signed by 15% of the legal voters (whose number is that of the total votes cast for governor, &c., as above) of a majority of the whole number of counties, but that in such case the law to be referred shall be inoperative until it is passed at the popular election. - If w is the weight of a locomotive in tons, r the radius of curvature of the track, v the velocity in feet per
**sec**.; then the horizontal force exerted on the bridge is wv 2 /gr tons. - According to modern measurements the solar radiation imparts almost 3 grammecalories of energy per minute per square centimetre at the distance of the earth, which is about Iï¿½3 X io s ergs per
**sec**. per cm. - Let E be the effective elasticity of the aether; then E = pc t, where p is its density, and c the velocity of light which is 3 X 10 10 cm./
**sec**. If = A cos" (t - x/c) is the linear vibration, the stress is E dE/dx; and the total energy, which is twice the kinetic energy Zp(d/dt) 2 dx, is 2pn2A2 per cm., which is thus equal to 1.8 ergs as above. - - Determine the time t
**sec**. and distance s ft. - Now, since v
**sec**i (54) di**sec**i dq C f(q**sec**i)' and multiplying by /dt or q, (55) dx C q**sec**i dq - f (q**sec**i)' and multiplying by dy/dx or tan i, (56) dy C q**sec**i tan dq - f (q**sec**i) ' also (57) di Cg dq g**sec**i .f (g**sec**i)' (58) d tan i C g**sec**i dq - q. - F (q
**sec**i)' from which the values of t, x, y, i, and tan i are given by integration with respect to q, when**sec**i is given as a function of q by means of (51). - Replacing then the angle i on the right-hand side of equations (54) - (56) by some mean value, t, we introduce Siacci's pseudovelocity u defined by (59) u = q
**sec**, t, so that u is a quasi-component parallel to the mean direction of the tangent, say the direction of the chord of the arc. - Di g d tan i g dt - v cos i ' and now (53) dx d 2 y dy d2xdx Cif dt 2 dt dt2 _ - _ gdt' and this, in conjunction with (46) dy _ d y tan i = dx dt/dt' (47)di d 2 d d 2 x dx
**sec**2 idt = (ctt d t - at dt2) I (dt), reduces to (48) Integrating from any initial pseudo-velocity U, (60) du t _ C U uf(u) x= C cos n f u (u) y=C sin n ff (a); and supposing the inclination i to change from 0, to 8 radians over the arc. - - tan B=C
**sec**n [I(U) - I (u)], while, expressed in degrees, (73) 0°-8° =C cos n [D(U) - D(u)]. - The difficulty is avoided by the use of Siacci's altitude-function A or A(u), by which y/x can be calculated without introducing sin n or tan n, but in which n occurs only in the form cos n or
**sec**n, which varies very slowly for moderate values of n, so that n need not be calculated with any great regard for accuracy, the arithmetic mean 1(0+0) of ¢ and B being near enough for n over any arc 4)-8 of moderate extent. - And 46
**Sec**. Of Mean Solar Time. - The self-conjugate circle is a 2 sin 2A +0 2 sin 2 B +y 2 sin 2C = o, or the equivalent form a cosAa 2 +bcosB(2 +ccosCy 2 = o, the centre being
**sec**A,**sec**B,**sec**C. - Xv.), which is held by some to be spurious, while others assign it to one or other of the years 287, 290, 296, 308 (so Mason, The Per
**sec**. of Diocl., pp. 275 seq.). - The number adopted for the value of the normal acceleration of gravity is 980.965 cm/
**sec-squared**. **Sec**. Hebr.- Now taking equation (72), and replacing tan B, as a variable final tangent of an angle, by tan i or dyldx, (75) tan 4) - dam= C
**sec**n [I(U) - I(u)], and integrating with respect to x over the arc considered, (76) x tan 4, - y = C**sec**n (U) - f :I(u)dx] 0 But f (u)dx= f 1(u) du = C cos n f x I (u) u du g f() =C cos n [A(U) - A(u)] in Siacci's notation; so that the altitude-function A must be calculated by summation from the finite difference AA, where (78) AA = I (u) 9 = I (u) or else by an integration when it is legitimate to assume that f(v) =v m lk in an interval of velocity in which m may be supposed constant. - Now calculate the pseudo-velocity uo from =v 95 cos 4)
**sec**n, and then, from the given values of 0 and 8, calculate u e from either of the formulae of (72) or (73): (82) I (u 9) - I (u0) tan 0 - tan 8 C**sec**n (83) D(ue) =D (uq5) 4)°-B° cos n' Then with the suffix notation to denote the beginning and end of the arc 0-0, mt e = C[Tum) - T (u0)], 5 ((x x9 1l 0. - - C = C cos n [5 (u 5) - S(ue)], ' y / e 0 A =tan - C
**sec**n [I (u 0) - S] A now denoting any finite tabular difference of the function between the initial and final (pseudo-) velocity. - Also the velocity v at the end of the arc is given by (87) ve = u e
**sec**0 cos n. - Gun to a shot weighing 380 lb fired with velocity 2375 f/s at elevation 40°; the range was about 12 m., with a time for flight of about 64
**sec**., shown in fig. - In direct fire the pseudo-velocities U and u, and the real velocities V and v, are undistinguishable, and
**sec**n may be replaced by unity so that, putting y =o in (79), (88) tan 4) = C [I (V) - y-s] Also (89) tan 4 - tan S=C [I(V) - L(v)] so that (9 °) tan 1 3=C [1 -s.