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916. Let L be the latitude of the place ; then from the right-angled A (L, 750) we have (by the question)

tan. L = cos. 4 x 150 x tan. 75o

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.. tan. L =

(2+3) 1 x

.(2+W3)

V(2-73) which indicates an error in the enunciation.

917. If p be the radius-vector of the comet’s orbit, and r its perihelion distance; then the time of its distance (from the sun) increasing from r to pis (see 484)

Xw (p-1) X (p + 2r)

3u

А. being the mass of the sun and comet.
In this formula let the mean distance of the earth from

P Р
the sun. Then, by the question,
2

2a
xva X (a +
3ye

3
10a

! 9

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Again, if T denote the length of the year; then (see 453), we

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хт

5 and t =

9 Xa and the number of days in which the comet is moving within the earth's orbit, is, therefore, expressed by

X T, 9 XT

10

2t

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MISCELLANIES.

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F =

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.

918. By 436, Vol. II. we have

dp
ca

d

2

de Now in the logarithmic spiral

0 = log. ?

1 .. do =

de da

3 dp

$

l.a
do
der

(la)
dla
ca

(la):
+
p3
c. {1 + (la)"} 1

F =

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919. The curve required is evidently the orthographic projection of the circular arch upon the surface of the water, and is, therefore, by the theory of projections, an ellipse, whose equation is

y = cot.' I . (72 – x2) and semi-axes

r and r cot. I. r being the radius of the arch, and I the meridian altitude of the

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Again the force at A being constant, we have

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a,

and udu =

urde
= 2ur (r = p)
:: 0 ::: v pole – go : v2r(

rs). But if r be the radius of a circle, and

S

cosine of an arc OC; then

pa = sin. and we also have
ch. a = ✓sin.' a + vers.'a

= 1 m2 5+(-3)*
= V272 2re

= N 2r . (r – s). Hence

iú :: sin. a : ch. a. See also Principia.

a.

921. The right-ascension of the star, and time of the year, being given, the angular distances of the star and sun from Aries are known, and, consequently, the angle between them. Let this be denoted by Also the co-declinations of the star and sun are known, which call D, and D' respectively. Then if L be the latitude of the place or altitude of the Pole; from the right-angled spherical triangles, whose hypothenuses are D, D', we easily get two equations between the two unknown quantities, L and the angular distance of the star from the meridian.

922. At any distance x from the centre of the sphere let the density be

Axe, being variable, and A and a certain constants not yet determined. Then if Q denote the quantity of matter in the sphere, whose radius is x, and M its magnitude, we have

4

76X)

d. Q = Ax x dM = Ax.de

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= 4 A7x+idx

4Απ
.. Q=

Os +3
Q being nothing when x = 0.

Now if D be the mean density of a sphere whose radius is r, then by the question mD = the density at its surface; and we have, from the quantities of matter being the same in both hypotheses of density,

4A7 Tpu3 =

@+3 But mD = Aro. 4Aart3 3m

a +3 .. 3m =a + 3

and a = 3. (m – 1). Hence Density = A.LU (-1)

a23(m-1),

D.

3

4A7

.

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923. Let G be the magnitude of a guinea; bG represents its weight C. Also let P and G be the magnitudes of the platina and silver requisite to make the coin in question; tben by the question we have

P +8= G
and ap + CS = G.
i. aP + aS = AG
and (c - a) S = (b a) G

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