Mathematical and Physical Papers, Volume 2University Press, 1884 - Mathematics |
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Page 6
... ( obtained by multiplying it by Joule's equivalent , 1390 ) is 83.4 × 2 95,000,000 441,000 = 386,900 ft . lbs . Now if , as Mr Waterston supposes , a meteor either strikes the Sun , or enters an atmosphere where the luminous and thermal ...
... ( obtained by multiplying it by Joule's equivalent , 1390 ) is 83.4 × 2 95,000,000 441,000 = 386,900 ft . lbs . Now if , as Mr Waterston supposes , a meteor either strikes the Sun , or enters an atmosphere where the luminous and thermal ...
Page 13
... obtained by meteors coming in thus directly from extra - planetary space . In conclusion , then , the source of energy from which solar heat is derived is undoubtedly meteoric . It is not any intrinsic energy in the meteors themselves ...
... obtained by meteors coming in thus directly from extra - planetary space . In conclusion , then , the source of energy from which solar heat is derived is undoubtedly meteoric . It is not any intrinsic energy in the meteors themselves ...
Page 50
... obtained , correspond- ing to ( 30 ) , in three dimensions , is interesting to pure mathema- ticians . XVI . Harmonic solutions . Any distribution of heat , whether in an infinite or a bounded solid , which keeps its type unchanged in ...
... obtained , correspond- ing to ( 30 ) , in three dimensions , is interesting to pure mathema- ticians . XVI . Harmonic solutions . Any distribution of heat , whether in an infinite or a bounded solid , which keeps its type unchanged in ...
Page 58
... obtain 8 92 € ̄x2 cos qxdx = π3 € ̄ 4 a [ * $$ ( a ) cos imx α S 12T2 с COS dx = πe α2 , when i is odd . " . ( 6 ) . The value of the first member is , of course , zero when i is even , because of the relation ( 4 ) ; and we conclude 2π ...
... obtain 8 92 € ̄x2 cos qxdx = π3 € ̄ 4 a [ * $$ ( a ) cos imx α S 12T2 с COS dx = πe α2 , when i is odd . " . ( 6 ) . The value of the first member is , of course , zero when i is even , because of the relation ( 4 ) ; and we conclude 2π ...
Page 59
... obtained by taking e = ( 1 ) , which gives a half its former value , so that we have 2π 3.2735 9 { ( -1 ) * % πTOC 10 COS + } ( - 1 ) 20 cos 10 3.2735 3πx 3.2735 + 25 5πα § ( · 1 ) 00 cos + & c . 3.2735 } ; of which three terms are ...
... obtained by taking e = ( 1 ) , which gives a half its former value , so that we have 2π 3.2735 9 { ( -1 ) * % πTOC 10 COS + } ( - 1 ) 20 cos 10 3.2735 3πx 3.2735 + 25 5πα § ( · 1 ) 00 cos + & c . 3.2735 } ; of which three terms are ...
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