Solutions of the problems and riders proposed in the Senate-house examination for 1854, by the moderators and examiners |
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Page 2
... passes through G , and , since OG is perpendicular to DE , DE touches it . = 2. 0 , A , B , C , are four points arranged in order in a straight line , so that OA , OB , OC , form an harmonic pro- gression . Prove that , A and C being ...
... passes through G , and , since OG is perpendicular to DE , DE touches it . = 2. 0 , A , B , C , are four points arranged in order in a straight line , so that OA , OB , OC , form an harmonic pro- gression . Prove that , A and C being ...
Page 13
... passing through the luminous point and the line . Therefore the two shadows on the ceiling are the intersections of the ceiling by the two planes passing through the longer candle and the two flames respectively ; the shadows if pro ...
... passing through the luminous point and the line . Therefore the two shadows on the ceiling are the intersections of the ceiling by the two planes passing through the longer candle and the two flames respectively ; the shadows if pro ...
Page 15
... passes through a rect- angular aperture in the floor : ascertain the form and area of the luminous patch on the floor of the room below . Shew that neither the shape nor the area of the patch will be affected by any movement of the disk ...
... passes through a rect- angular aperture in the floor : ascertain the form and area of the luminous patch on the floor of the room below . Shew that neither the shape nor the area of the patch will be affected by any movement of the disk ...
Page 11
... passes through the given point . 9. The equation for the projection of the Moon's radius vector on the ecliptic is P and h2u2 h2 u2 = Ꮭ ᎢᏧᎾ hus deu P T du d2u + u = 2 d02 + u h2u2 h2u3 do do 8 1 + 3 cos 2 ( 0 - 0 ' ) 2 # ( 1 - 30 ...
... passes through the given point . 9. The equation for the projection of the Moon's radius vector on the ecliptic is P and h2u2 h2 u2 = Ꮭ ᎢᏧᎾ hus deu P T du d2u + u = 2 d02 + u h2u2 h2u3 do do 8 1 + 3 cos 2 ( 0 - 0 ' ) 2 # ( 1 - 30 ...
Page 13
... passing through the luminous point and the line . Therefore the two shadows on the ceiling are the intersections of the ceiling by the two planes passing through the longer candle and the two flames respectively ; the shadows if pro ...
... passing through the luminous point and the line . Therefore the two shadows on the ceiling are the intersections of the ceiling by the two planes passing through the longer candle and the two flames respectively ; the shadows if pro ...
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Solutions of the Problems and Riders Proposed in the Senate-House ... Exam Papers Cambridge Univ No preview available - 2016 |
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angular velocity asymptote axes Cambridge catenary centre of force centre of gravity chord circle cloth conic section constant cos² cose cosẞ cota cotß Crown 8vo curvature curve cylinder denoting described diameter direction distance dx dy dy dx ecliptic elastic ellipse equal equation equilibrium Fellow of St fixed point fluid geometrical progression given Hence horizontal hyperbola inclined plane intersection lamina latus rectum length locus longitude M.A. Fellow major axis middle point motion orbit parabola parallel particle passing perpendicular position pressure projection prove radius refraction right angles ring shew sides Similarly sine sino sinẞ straight line string Supposing surface tana tangent tanß triangle Trinity College tube V₁ vertical
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