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Ball bearings are rated by a manufacturer for a life of 106 revolutions. The catalogue rating of a particular bearing is 16 kN. If the design load is 2 kN, the life of the bearing will be p x 106 revolutions, where p is equal to _____. (Answer up to the nearest integer)
    Correct answer is '512'. Can you explain this answer?
    Most Upvoted Answer
    Ball bearings are rated by a manufacturer for a life of 106 revolutio...
    Given information:
    - The manufacturer rates the life of the ball bearing as 106 revolutions.
    - The catalogue rating of the bearing is 16 kN.
    - The design load is 2 kN.

    To calculate the life of the bearing, we need to determine the ratio of the catalogue rating to the design load and then multiply it by the rated life of the bearing.

    Ratio of Catalogue Rating to Design Load:
    - The ratio is calculated as the catalogue rating divided by the design load.
    - In this case, the catalogue rating is 16 kN and the design load is 2 kN.
    - Therefore, the ratio is 16 kN / 2 kN = 8.

    Life of the Bearing:
    - To find the life of the bearing, we multiply the ratio by the rated life of the bearing.
    - The rated life of the bearing is given as 106 revolutions.
    - Therefore, the life of the bearing is 8 x 106 revolutions = 8,000,000 revolutions.

    Final Answer:
    - The value of p, which represents the life of the bearing, is 8,000,000 revolutions.
    - However, we need to round this value to the nearest integer as per the question's instructions.
    - Rounding 8,000,000 to the nearest integer gives us 8,000,000, which is equal to 512 x 106 revolutions.

    Therefore, the answer is 512.
    Free Test
    Community Answer
    Ball bearings are rated by a manufacturer for a life of 106 revolutio...
    Given that
    Catalogue Rating = 16 kN
    Design load = 2 kN
    We have to find the life of bearing = ?
    We know that
    L = (C/F)q million revolution
    Where C = Dynamic load capacity
    F = Equivalent dynamic load
    q = 3 For ball bearing
    So C = 16 kN, F= 2 kN
    L = [16/2]3 = [8]3 = 512 million revolution
    Life of bearing = p X 106 revolution
    = 512 X 106 revolution
    So p = 512
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    Ball bearings are rated by a manufacturer for a life of 106 revolutions. The catalogue rating of a particular bearing is 16 kN. If the design load is 2 kN, the life of the bearing will be p x 106 revolutions, where p is equal to _____. (Answer up to the nearest integer)Correct answer is '512'. Can you explain this answer?
    Question Description
    Ball bearings are rated by a manufacturer for a life of 106 revolutions. The catalogue rating of a particular bearing is 16 kN. If the design load is 2 kN, the life of the bearing will be p x 106 revolutions, where p is equal to _____. (Answer up to the nearest integer)Correct answer is '512'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about Ball bearings are rated by a manufacturer for a life of 106 revolutions. The catalogue rating of a particular bearing is 16 kN. If the design load is 2 kN, the life of the bearing will be p x 106 revolutions, where p is equal to _____. (Answer up to the nearest integer)Correct answer is '512'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Ball bearings are rated by a manufacturer for a life of 106 revolutions. The catalogue rating of a particular bearing is 16 kN. If the design load is 2 kN, the life of the bearing will be p x 106 revolutions, where p is equal to _____. (Answer up to the nearest integer)Correct answer is '512'. Can you explain this answer?.
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