A student solving a physics problem to find the unknown has applied physics principles and obtained the expression: μkmgcosθ=mgsinθ−ma, wher

Question

A student solving a physics problem to find the unknown has applied physics principles and obtained the expression: μkmgcosθ=mgsinθ−ma, where g=9.80meter/second2, a=3.60meter/second2, θ=27.0∘, and m is not given. Which of the following represents a simplified expression for μk?A student solving a physics problem to find the unknown has applied physics principles and obtained the expression: , where , , , and is not given. Which of the following represents a simplified expression for ?tanθ− agTo avoid making mistakes, the expression should not be simplified until the numerical values are substituted.gsinθ−agcosθThe single equation has two unknowns and cannot be solved with the information given.

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Latifah 6 months 2021-07-19T04:32:28+00:00 1 Answers 106 views 0

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    2021-07-19T04:34:04+00:00

    Solution :

    Given expression :

    $\mu_k$mgcosθ = mgsinθ − ma

    Here, g = 9.8 m/s^2 , a = 3.60 m/s^2 , θ = 27°

    Therefore,

    $\mu_k mg \cos \theta = mg \sin \theta - ma$

    $\mu_k mg \cos \theta = m(g \sin \theta - a)$

    $\mu_k g \cos \theta = (g \sin \theta - a)$

    $\mu_k =\frac{(g \sin \theta-a)}{g \cos \theta}$

    Mow calculating the coefficient of kinetic friction as follows :

    $\mu_k=\frac{g \sin \theta-a}{g \cos \theta}$

    $\mu_k=\frac{9.8 \times  \sin 27^\circ-3.60}{9.8 \times \cos 27^\circ}$

    $\mu_k=0.097$

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