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Solutions of the Problems and Riders Proposed in the Senate-House Examination for 1854, by the Moderators and Examiners (Paperback) Loot Price: R312
Discovery Miles 3 120
Solutions of the Problems and Riders Proposed in the Senate-House Examination for 1854, by the Moderators and Examiners...

Solutions of the Problems and Riders Proposed in the Senate-House Examination for 1854, by the Moderators and Examiners (Paperback)

Exam Papers Cambridge Univ

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Loot Price R312 Discovery Miles 3 120

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This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1854 Excerpt: ...from the focus S on the tangent at P. Produce it to Q, so that SQ bears a certain ratio to the velocity at P, and in the axis take a point B, such that SB bears the same ratio to the velocity at the vertex A; and join BQ. Then SQ cc velocity at P cc-y., or SY. SQ = a constant quantity = SA. SB, or SY: SA:: SB: SQ. And the triangles ASY, QSB, having a common angle Q at S, and their sides about that angle proportional, are similar. Hence the angle BQS = the angle SAY = a right angle, and Q will always lie on the circle whose diameter is SB. 4. Given the velocities and the directions of motion at any three points of an orbit described under the action of a central force: find the centre of force. If the velocities at the three points be respectively parallel and proportional to the opposite sides of the triangle of which they are the angular points, the centre of force is the centre of gravity of the triangle. Let P, Q, B (fig. 77) be three points of a central orbit, at each of which the velocity is parallel and proportional to the opposite side of the triangle PQB: produce the tangents at P, Q, B so as to form a new triangle PQ'B', having its sides parallel and proportional to those of PQB. Join PP: because the perpendiculars from the centre of force on PQ', PB' are inversely proportional to the velocities at B, Q, they are inversely proportional to the sides P Q', PB'; therefore the triangles, whose common vertex is the centre of force, and whose bases are the sides PQ PB will be equal, and therefore the centre of force will lie in the line PP: so also it will lie in the line QQ, and will be the centre of gravity of the triangle PQB, for the lines PP, QQ' bisect respectively the sides QB, BP. K2 5. An ellipse is described under the action of a force tending ..

General

Imprint: Rarebooksclub.com
Country of origin: United States
Release date: March 2012
First published: March 2012
Authors: Exam Papers Cambridge Univ
Dimensions: 246 x 189 x 2mm (L x W x T)
Format: Paperback - Trade
Pages: 44
ISBN-13: 978-1-130-18965-0
Categories: Books > Humanities > History > General
Books > History > General
LSN: 1-130-18965-1
Barcode: 9781130189650

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