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 78
... inch , occupies n cubic inches when the temperature is t ; find how many cubic inches it will occupy under the pressure of m ' pounds to the square inch when the temperature is t ' . The answer to the second part of ( A ) is 1 + at II ...
... inch , occupies n cubic inches when the temperature is t ; find how many cubic inches it will occupy under the pressure of m ' pounds to the square inch when the temperature is t ' . The answer to the second part of ( A ) is 1 + at II ...
Side 80
... inches . Let h , x , be respectively the true and false readings of the barometer ; c the length of tube occupied by ... inches ; and therefore the error produced is 2 inches . Also let c = c1 ; therefore c..29.2 • 2 1 ; a C1 29.2 ...
... inches . Let h , x , be respectively the true and false readings of the barometer ; c the length of tube occupied by ... inches ; and therefore the error produced is 2 inches . Also let c = c1 ; therefore c..29.2 • 2 1 ; a C1 29.2 ...
Side 81
... inches , so that therefore 1850 . Let c 30.3 = 1.01 ; C2 30 1.6 = 1 · 01 . 101 ⚫016 of an inch . ( A ) . Describe Nicholson's Hydrometer . ( B ) . Given two weights which cause the instrument to sink to a certain depth in two fluids ...
... inches , so that therefore 1850 . Let c 30.3 = 1.01 ; C2 30 1.6 = 1 · 01 . 101 ⚫016 of an inch . ( A ) . Describe Nicholson's Hydrometer . ( B ) . Given two weights which cause the instrument to sink to a certain depth in two fluids ...
Side 89
... The least distance between an object and its image formed by a plano - convex lens of glass is 12 inches ; the index of refraction being , find the radius of the spherical surface . The focal length of a convex lens may be found OPTICS .
... The least distance between an object and its image formed by a plano - convex lens of glass is 12 inches ; the index of refraction being , find the radius of the spherical surface . The focal length of a convex lens may be found OPTICS .
Side 90
... inches ; therefore f = 23 inches . 4 But if r be the radius of the curved surface , 1 1 ) . — , disregarding the sign , 1 ( μ - 1 ) f ( 3 - 1 ) 1 ) 1 ; 2r therefore 1 зо If in . 1848 . ( A ) . Describe the Astronomical Telescope 90 ...
... inches ; therefore f = 23 inches . 4 But if r be the radius of the curved surface , 1 1 ) . — , disregarding the sign , 1 ( μ - 1 ) f ( 3 - 1 ) 1 ) 1 ; 2r therefore 1 зо If in . 1848 . ( A ) . Describe the Astronomical Telescope 90 ...
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.