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1 year ago

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4. A 20-kg box sits on an incline of 30° from the horizontal. The coefficient of friction between the box and the incline is o.3

0. Find the acceleration of the box down the incline.

1 answer:

Slav-nsk [51]1 year ago

6 0

The box accelerates down the incline at 2.4m/s^2.

<h3>How do you find the acceleration of an incline box?</h3>

A mass m particle will slide down a smooth inclined plane if it is released onto it with a friction force F=0. We resolve in the direction of motion to determine the particle's acceleration as it slides. F = ma, mg cos(90) = ma, gg cos(90) = a, gg sin() = a.

By using zigmaFx = max, we will determine the acceleration.

However, we must first determine the friction force Ff.

Because cos 30 degree = 0.866 and Fy = ma y = 0 result in FN - 0.87mg = 0, we may calculate FN as (0.87)(20Kg)(9.81 m/s2) = 171N.

From Ff = mue FN = 0.30)(171 N)= 51N, we can now calculate Ff.

We get Ff - 0.50mg = ma x 51N - (0.50)(20)(9.81)N = (20kg)(ax) from the expression zigmaFx = max, where ax = -2.35 m/s2.

At 2.4 m/s2, the box quickens its descent of the hill.

To learn more about acceleration refer to:

brainly.com/question/460763

#SPJ1

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B

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From the question we are told that

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From the question we are told that

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The order of the fringe is $n = 2$

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Generally the slit width is mathematically represented as

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Generally the condition for constructive interference for the first ray is mathematically represented as

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=> $\theta_1 = sin^{-1} [\frac{ 2 * \lambda }{d}]$

=> $\theta_1 = sin^{-1} [\frac{ 2 * 650 *10^{-9} }{ 2*10^{-6}}]$

=> $\theta_1 = 40.5 ^o$

Generally the condition for constructive interference for the second ray is mathematically represented as

$d sin(\theta_2) = n * \lambda_2$

=> $\theta_2 = sin^{-1} [\frac{ 2 * \lambda_1 }{d}]$

=> $\theta_2 = sin^{-1} [\frac{ 2 * 420 *10^{-9} }{ 2*10^{-6}}]$

=> $\theta_2 = 24.8 ^o$

Generally the angular separation is mathematically represented as

$\theta = \theta_1 - \theta_1$

=> $\theta = 42.5^o - 24.8^o$

=> $\theta = 15.7^o$

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