Solutions to the mathematical examination papers set for admission to the Royal military academy, Woolwich, and for the Royal military college [&c.] by D. Tierney and H. Sharratt1877 |
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Página 33
... velocity be that with which a body passes over a feet in t seconds , and the unit of acceleration that which produces in t seconds a velocity of b feet per t seconds , find the units of space and time . Parkinson's Mechanics , Arts . 5 ...
... velocity be that with which a body passes over a feet in t seconds , and the unit of acceleration that which produces in t seconds a velocity of b feet per t seconds , find the units of space and time . Parkinson's Mechanics , Arts . 5 ...
Página 34
... velocity , velocity and space in the case of a heavy body falling freely in vacuo . Parkinson's Mechanics , Arts . 67 , 68 , 69 , 70 . If s , be the space described in t seconds by a body projected directly towards a centre of uniform ...
... velocity , velocity and space in the case of a heavy body falling freely in vacuo . Parkinson's Mechanics , Arts . 67 , 68 , 69 , 70 . If s , be the space described in t seconds by a body projected directly towards a centre of uniform ...
Página 35
... velocity . Parkinson's Mechanics , Art . 75 . Shew that the acceleration of the common centre of gravity of the particles is g ( ་ P - q \ p + q ) 2 Let 0 ( fig . 5 ) be the point of bisection of the line joining p and q , then this ...
... velocity . Parkinson's Mechanics , Art . 75 . Shew that the acceleration of the common centre of gravity of the particles is g ( ་ P - q \ p + q ) 2 Let 0 ( fig . 5 ) be the point of bisection of the line joining p and q , then this ...
Página 36
... velocity at the vertex of the parabola , prove that the time of describing the V1v sina arc = до 2 Let 0 ,, 0 , be the inclination to the horizon of the particles ' direction of motion at the two places respec- tively . Then we have and ...
... velocity at the vertex of the parabola , prove that the time of describing the V1v sina arc = до 2 Let 0 ,, 0 , be the inclination to the horizon of the particles ' direction of motion at the two places respec- tively . Then we have and ...
Página 37
... velocity equal to that due to the height EX , then the particle will , after rebounding from the wall Bb , · strike the particle at D. A similar method of construction will determine the velocity and direction of projection so that the ...
... velocity equal to that due to the height EX , then the particle will , after rebounding from the wall Bb , · strike the particle at D. A similar method of construction will determine the velocity and direction of projection so that the ...
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Solutions To The Mathematical Examination Papers Set For Admission To The ... D Tierney,Handell Sharratt Sin vista previa disponible - 2023 |
Solutions to the Mathematical Examination Papers Set for Admission to the ... D. Tierney,Handell Sharratt Sin vista previa disponible - 2016 |
Solutions To The Mathematical Examination Papers Set For Admission To The ... D. Tierney,Handell Sharratt Sin vista previa disponible - 2023 |
Términos y frases comunes
AB² ABCD acceleration AD² ARITHMETIC axis BC² cent centre of gravity circle coefficient of friction Conic Sections cose cubic curve decimal described diameter Differential Calculus directrix Divide dy dx equal and parallel Euclid expression feet Find the equation forces fraction given straight line hyperbola inches inclined integration intersect Join latus rectum least common multiple logarithmic mechanical advantage METCALFE AND SON moment of inertia Multiply opposite angles parabola parallelogram Parkinson's Mechanics particle perpendicular plane point of bisection Prop prove quadrilateral radius ratio rectangle contained Result right angles roots seconds segments semicircle shew sides Similarly sin² sine square string subtending Subtract tangent Todhunter Todhunter's Trigonometry triangle ABC Trig vertex vertical virtual velocities weight whence whole number yards
Pasajes populares
Página 55 - If two triangles have two sides of the one equal to two sides of the...
Página 71 - If a straight line be divided into two equal parts, and also into two unequal parts, the rectangle contained by the unequal parts, together with the square on the line between the points of section, is equal to the square on half the line.
Página 11 - ... shall be equal to three given straight lines, but any two whatever of these must be greater than the third.
Página 12 - 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.
Página 13 - 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.
Página 15 - Similar triangles are to one another in the duplicate ratio of their homologous sides.
Página 13 - BAC is cut off from the given circle ABC containing an angle equal to the given angle D : Which was to be done. PROP. XXXV. THEOR. If two straight lines within a circle cut one another, the rectangle contained by the segments of one of them is equal to the rectangle contained by the segments of the other.
Página 62 - In every triangle, the square of the side subtending either of the acute angles, is less than the squares of the sides containing that angle, by twice the rectangle contained by either of these sides, and the straight line intercepted between the acute angle and the perpendicular let fall upon it from the opposite angle.
Página 13 - PROP. X. THEOR. IF a straight line be bisected, and produced to any point, the square of the whole line thus produced, and the square of the part of it produced, are together double of the square of half the line bisected, and of the square of the line made up of the half and the part produced.
Página 70 - And because the angle ABC is equal to the angle BCD, and the angle CBD to the angle ACB, therefore the whole angle ABD is equal to the whole angle ACD • (ax.