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 65
... cycloid , the time of an oscillation will be independent of the arc of vibration . a2 2 ( B ) . A seconds pendulum ... cycloid . If a pendulum of length 7 make small oscillations about one extremity , we may consider the other extremity ...
... cycloid , the time of an oscillation will be independent of the arc of vibration . a2 2 ( B ) . A seconds pendulum ... cycloid . If a pendulum of length 7 make small oscillations about one extremity , we may consider the other extremity ...
Side 69
... cycloid ? ( B ) . A pendulum which oscillates seconds at one place is carried to a place where it gains two minutes a - day ; compare the force of gravity at the latter place with that at the former . When a body oscillates in a cycloid ...
... cycloid ? ( B ) . A pendulum which oscillates seconds at one place is carried to a place where it gains two minutes a - day ; compare the force of gravity at the latter place with that at the former . When a body oscillates in a cycloid ...
Side 70
... cycloid . The radius of curvature at C , ( fig . 50 ) , CP2 limit Pn limit ( 2PR ) " Pn 2 by ( A ) , 2 limit PC2 2 CB . pn " Again , by means of the property enunciated in ( A ) , it is proved that the time of oscillation in the cycloid ...
... cycloid . The radius of curvature at C , ( fig . 50 ) , CP2 limit Pn limit ( 2PR ) " Pn 2 by ( A ) , 2 limit PC2 2 CB . pn " Again , by means of the property enunciated in ( A ) , it is proved that the time of oscillation in the cycloid ...
Side 71
Francis James Jameson. of which CpB is the circle of curvature , is π = 1 / 1π CB 9 CB √ ( 20B ) = time of oscillation in the original cycloid . HYDROSTATICS . 1850. ( A ) . Find the pressure DYNAMICS . 71.
Francis James Jameson. of which CpB is the circle of curvature , is π = 1 / 1π CB 9 CB √ ( 20B ) = time of oscillation in the original cycloid . HYDROSTATICS . 1850. ( A ) . Find the pressure DYNAMICS . 71.
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The Principles of the Solution of the Senate-House 'Riders: Exemplified by ... Francis J. Jameson Ingen forhåndsvisning - 2018 |
The Principles of the Solution of the Senate-House 'Riders: Exemplified by ... Francis J. Jameson Ingen forhåndsvisning - 2015 |
Almindelige termer og sætninger
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 passager
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.