The Principles of the Solution of the Senate-house 'riders,' Exemplified by the Solution of Those Proposed in the Earlier Parts of the Examinations of the Years 1848-1851Macmillan & Company, 1851 - 116 sider |
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Side vi
... weight is attributed to it by the Moderators and Examiners , in estimating the relative merits of Candidates for Honours . ' A few observations therefore on the principles of the solution of this class of questions , exemplified by the ...
... weight is attributed to it by the Moderators and Examiners , in estimating the relative merits of Candidates for Honours . ' A few observations therefore on the principles of the solution of this class of questions , exemplified by the ...
Side 46
... weight acting along GW , tension of string along BP , reaction of wall along ARI NP . Now the principle of which ( 4 ) is the enunciation tells us that these three forces must meet in a point ( because they are not parallel ) in order ...
... weight acting along GW , tension of string along BP , reaction of wall along ARI NP . Now the principle of which ( 4 ) is the enunciation tells us that these three forces must meet in a point ( because they are not parallel ) in order ...
Side 48
... weight rests in equi- librium with its base on a horizontal plane sufficiently rough to prevent all sliding . A force acts upon it in its own plane and in a given line drawn through the vertex and without the triangle : find by a ...
... weight rests in equi- librium with its base on a horizontal plane sufficiently rough to prevent all sliding . A force acts upon it in its own plane and in a given line drawn through the vertex and without the triangle : find by a ...
Side 49
... weight of the triangle and the force which acts at the vertex shall cut the base of the triangle . Let the vertical through the centre of gravity G ( fig . 37 ) of the triangle meet the given line through A in O , and the base of the ...
... weight of the triangle and the force which acts at the vertex shall cut the base of the triangle . Let the vertical through the centre of gravity G ( fig . 37 ) of the triangle meet the given line through A in O , and the base of the ...
Side 51
... weight may be supposed to act , G may be considered as the point of application of the resultant of the weights of the portions Abc , bc CB . Hence therefore Now GG ' : Gg : weight of Abc : weight of bc CB , :: area Abc : area bc CB ...
... weight may be supposed to act , G may be considered as the point of application of the resultant of the weights of the portions Abc , bc CB . Hence therefore Now GG ' : Gg : weight of Abc : weight of bc CB , :: area Abc : area bc CB ...
Andre utgaver - Vis alle
The Principles of the Solution of Senate-house 'riders': Exemplified by the ... Francis James Jameson Uten tilgangsbegrensning - 1851 |
The Principles of the Solution of the Senate-House 'Riders: Exemplified by ... Francis J. Jameson Ingen forhåndsvisning tilgjengelig - 2018 |
The Principles of the Solution of the Senate-House 'Riders: Exemplified by ... Francis J. Jameson Ingen forhåndsvisning tilgjengelig - 2015 |
Vanlige uttrykk og setninger
AC² AN.NM Arithmetic arithmetical progression axis bisects body C₁ Cambridge centre of gravity chord CHURCHILL BABINGTON circle cloth cone Conic Sections conjugate hyperbola constant curvature curve cycloid describe diameter direction directrix distance drawn Edition ellipse equations equilibrium Fellow of St fluid focus geometrical given point Hence horizontal hyperbola inches inclined inscribed John's College joining latus-rectum least common multiple Lemma length locus meet mirror move number of seconds oscillation parabola parallel parallelogram particle perpendicular plane polygon pressure prop proportional proposition prove pullies quadrilateral quantity radius ratio rays rectangle refraction right angles sewed shew sides specific gravity spherical square straight line string surface tan² tangent triangle ABC Trinity College tube V₁ vary vertex vertical W₁ weight
Populære avsnitt
Side 4 - To divide a given straight line into two parts, so that the rectangle contained by the whole and one of the parts, shall be equal to the square of the other part.
Side 6 - The angle at the centre of a circle is double of the angle at the circumference upon the same base, that is, upon the same part of the circumference.
Side 11 - AB is a diameter, and P any point in the circumference of a circle; AP and BP are joined and produced if necessary ; if from any point C of AB, a perpendicular be drawn to AB meeting AP and .BP in points D and E respectively, and the circumference of the circle in a point F, shew that CD is a third proportional of CE and CF.
Side 9 - IF the angle of a triangle be divided into two equal angles, by a straight line which also cuts the base; the segments of the base shall have the same ratio which the other sides of the triangle have to one another...
Side 4 - In any right-angled triangle, the square which is described upon the side subtending the right angle, is equal to the squares described upon the sides which contain the right angle.