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Problems in Mathematical Analysis.pdf - pwp.net.ipl.pt

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_Answers_433<br />

The force of <strong>in</strong>teraction of charges is / 7 = ^i dynes. Consequently, the work<br />

Jd.x<br />

-J-=S<br />

*'<br />

/ 1 1 \<br />

= --- = *o*i(<br />

1.8-10<br />

) 4<br />

ergs. 1762. 4=800 nln2 kgm. Solution. For an<br />

* X X \ 2 /<br />

isothermal process, pw = p tV The work performed <strong>in</strong> the expansion of a gas<br />

from volume v to<br />

v\<br />

volume v is A= \<br />

l<br />

p di>=p i> In ^ . 1763.<br />

J ^0<br />

A =5= 15,000 kgm.<br />

Solution. For an adiabatic<br />

UQ<br />

process, the Poisson law pv k = t pj>^ where<br />

k<br />

*^1.4, holds true. Hence A = ( P -^dv- -Ms<br />

J v<br />

k k I<br />

4<br />

1764. 4=~jiu.Pa. Solution. If a is the radius of the base of a shaft, them<br />

p<br />

the pressure on unit area of the support p = z . The frictional force of a<br />

3ta a<br />

2uP<br />

r<strong>in</strong>g of width dr, at a distance r from the centre, is -^~rdr. The work per-<br />

formed by frictional forces on a r<strong>in</strong>g <strong>in</strong> one complete revolution is<br />

i>i<br />

. Therefore, the complete work A=^x f r* dr = JifiPa,<br />

1765. j M# 2 r<br />

o b<br />

1770. P = abyxh. 1771. P = ^-_ (the vertical component is directed upwards).<br />

1772. 633 Igm 1773. 99.8 cal. 1774. M^~ gf cm. 1775. (k<br />

gravitational constant). 1776.^-<br />

15-1900<br />

a a<br />

. Solution. Q= (v2nrdr = 2<br />

__ . n 2 a6 8<br />

UJ A<br />

1777 - Q== ^^-^ Hlnt -<br />

is the<br />

f(a r*)rdr

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