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Now since the fluids are elastic, they will always fill the vessels ; hence the velocity with which the denser air would rush into a vacuum at the end of this time is

V =

✓ 2gh x

x

D h being the altitude of a cylinder of water equivalent to the weight of the atmosphere.

But the moving forces with which this would take place, and with which the same fluid would rush into the vessel containing the rarer air, are respectively E

E
X D', and X (D' — d')

d

and these moving forces are as the squares of the velocities

E

E ..

(D - d') :: Vo : 0,2

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D:

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Again, if C, c, denote the capacities of the vessels, we have
CXD+cxd=CD + cx d'

С
' Ꭰ . (D – D') - d

-
:: v8 = 2ghd x

D'

с

C + C. D'- cd + CD = 2gh x

cD'2 which gives the velocity for any degree of density of the denser medium. If D' be eliminated, a similar formula will be obtained, giving the velocity corresponding to any density of the rarer medium.

761. Generally, supposing the basin and tube cylindrical, let R, r, be their radii in inches, and when the mercury in the tube shall have descended x inches, let that in the basin ascend through X inches; then

qora x x = (TR’ — r*) X

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Hence the real depression of the mercury is

RPC
R- p2

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762. The spaces decrease in Harmonical Progression, (Vince, Prop. LXX). Hence if x be the space after the first descent, we have (by the question and the nature of the progression)

12 : 6 :: 12 X : - 6
.. 18x = 2 X 72

.. X = 8

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1

is the common difference of the re12 8

24 procals of the spaces, so that the nch

space

is 1

24 1 =

12 24 n+2

n

763. Let x', a' be the depressions after immersion, the standard altitude being then 30 inches. Then (749) and by the question

22 + (a -- 30) x = 306

}

.

whence

- (a – 30) + n la – 30)* + 1206

2

fa' – 30) + V (a' - 30)2 + 1206'

A'

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A

Again, let h be the new standard altitude, then we have

(x + 1) + (a - b) (x + 1) = hb

(.x + 1)' + a. (x + 1) A + 1 .. h =

X (A+a+1) b

b is known. Hence if y denote the corresponding depression in the second tube, we have (749)

- (d -- 1) + va'- h) + 4h6

2

Q.E.I.

764. The number of degrees in Fahrenheit's thermometer between the freezing and boiling points of water is

212 - 32 = 180. Hence if the diminution of bulk be Lth of the whole for each

n

of these degrees, 2r the diameter of the tube, and B the magnitude
of the bulb; then the several diminutions will be
212

2117ra
anra B +
180

180

&c.

B +

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n2

n

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765. Generally, let R be the radius of the sphere, , the radius of the tube; then the surface of the segment of the sphere is (Vince's Fluxions, p. 110,)

S = 4*R® – 2*R X 2Rr - p Moreover the centre of gravity of this surface is distant from the centre of the base of the segment by (Vince's Fluxions, p. 118,):

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Hence the pressure upon the surface of the ball is
P= Sx G + 10 x density of the fluid

= weight of a volume S x(G + 10) of the mercury. Now, when the mercury just fills the segment of the sphere, let the weight of a cubic inch of it be w; then since the volume of the segment of the sphere is (Vince's Fluxions, p. 95)

M= » (2R – V2Rr–r)*x (R 2R - N2Rr pe

3

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the weight of a cubic inch when the mercury has risen 10 inches in the tube is

M

M + 107r and the required pressure is finally expressed by

Mw

XS X(G + 10).
M + 1071%

P=

766. Let å be the density of the air at first, h the height of a homogenous atmosphere; D the density of the water; then the water would rush into the vessel, completely empty, with a velocity

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and with a moving force measured by

M = 8h + Da. and when the water shall have risen through the space x, the moving force will be

1 M' = dh + Da

4r3

4773

X

xdh

3

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3

= Da

q? (3r – x)

4r3 - x*(3r - x) But Moc square of the velocity. Hence the velocity at this stage of the filling, viz.

va

x*. (3r - x) Da

dh 13

4r3 v

{2ga.

x* (3r – x) M

8h + Da is known.

Let u be 0. Then at first the water rushes into the globe with the velocity

2gD

sh + Da Again, when v = 0, we have x". (3r – 2) = Da (4,8 — 3r – x)

4Dar3 :: ** (31 —- ) =

1+ Da Hence, the portion of the sphere which the water then occupies, is

40 x xP (3r — x) =

=

1+ Da By the resolution of a cubic equation x also may be found.

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ars

3

3

767. Let a be the altitude and r the radius of the base of the cylinder, and A the area of the orifice. Also let S be the density at first of the air, E its elasticity at first; s its density, and e its elasticity at the end of the time t. Also, let V and v denote the initial and subsequent velocities of the issuing medium.

Now, supposing Q R S the mass of air expelled when the piston has descended through a - x, we have

QXS = ane x aS - #r*x xs ·(1) and since the expelling force at first and afterwards are

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since the moving forces are as the quantities of motion generated in a given time.

Hence, v = V or it is constant.

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