Solutions of the problems and riders proposed in the Senate-house examination for 1854, by the moderators and examiners |
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Resultat 1-5 av 32
Side 10
... greatest compression is V - 2R ' cos 30 ° = V - R'√3 ; therefore the velocity of A , resolved along the normal AB , is ( V – R′√3 ) cos30 ° = √3 V – 4 R ' ; 2 V- and the velocity of B , resolved along the same line AB , is R ' ; but ...
... greatest compression is V - 2R ' cos 30 ° = V - R'√3 ; therefore the velocity of A , resolved along the normal AB , is ( V – R′√3 ) cos30 ° = √3 V – 4 R ' ; 2 V- and the velocity of B , resolved along the same line AB , is R ' ; but ...
Side 11
... greatest com- pression = √3 - V – R ' . Also the normal velocity of B before impact = V cos30 ° = √3 V , 2 therefore the normal velocity of B at the time of greatest com- pression = R ' √3 V ; 2 therefore , equating these normal ...
... greatest com- pression = √3 - V – R ' . Also the normal velocity of B before impact = V cos30 ° = √3 V , 2 therefore the normal velocity of B at the time of greatest com- pression = R ' √3 V ; 2 therefore , equating these normal ...
Side 10
... greatest compression is V2R ' cos 30 ° – V – R'√3 ; = therefore the velocity of A , resolved along the normal AB , is ( V – R'√ / 3 ) cos 30 ° = √3 v — § R ′ ; - V - and the velocity of B , resolved along the same line AB , is R ...
... greatest compression is V2R ' cos 30 ° – V – R'√3 ; = therefore the velocity of A , resolved along the normal AB , is ( V – R'√ / 3 ) cos 30 ° = √3 v — § R ′ ; - V - and the velocity of B , resolved along the same line AB , is R ...
Side 11
... greatest com- pression = √ / 3 - V — §R ' . - Also the normal velocity of B before impact = V cos 30 ° = V , 2 therefore the normal velocity of B at the time of greatest com- pression = R ' - V ; therefore , equating these normal ...
... greatest com- pression = √ / 3 - V — §R ' . - Also the normal velocity of B before impact = V cos 30 ° = V , 2 therefore the normal velocity of B at the time of greatest com- pression = R ' - V ; therefore , equating these normal ...
Side 23
... greatest possible ; therefore And from ( 1 ) P therefore α b = { = 0 = adp + bdq . 0 = pdp + qdq ; 1 √ ( a2 + b2 ) therefore the volume AOBG and = ap + bq ; a2 + b ab = 2 √√ ( a2 + b2 ) , DG = ap + bq = √ ( a + b2 ) = OD ; therefore ...
... greatest possible ; therefore And from ( 1 ) P therefore α b = { = 0 = adp + bdq . 0 = pdp + qdq ; 1 √ ( a2 + b2 ) therefore the volume AOBG and = ap + bq ; a2 + b ab = 2 √√ ( a2 + b2 ) , DG = ap + bq = √ ( a + b2 ) = OD ; therefore ...
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Solutions of the Problems and Riders Proposed in the Senate-House ... Exam Papers Cambridge Univ Ingen forhåndsvisning tilgjengelig - 2016 |
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Populære avsnitt
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