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Eduardwww [97]
3 years ago
13

A brick is resting on a smooth wooden board that is at a 30° angle. What is one way to overcome the static friction that is hold

ing the brick in place?Please help i will give brainliest!!!! lower the board so it's level with the ground ____ roughen up the texture of the wooden board ___ raise the board to a higher angle ____ press down on the brick in a direction that is perpendicular to the board _____
Physics
1 answer:
lubasha [3.4K]3 years ago
4 0

Answer:

to overcome the out of friction we must increase the angle of the plane

Explanation:

To answer this exercise, let's propose the solution of the problem, write Newton's second law. We define a coordinate system where the x axis is parallel to the plane and the other axis is perpendicular to the plane.

X axis

       fr - Wₓ = m    a                      (1)

Y axis  

       N- W_{y} = 0

       N = W_{y}

let's use trigonometry to find the components of the weight

        sin θ = Wₓ / W

        cos θ = W_{y} / W

        Wₓ = W sin θ

        W_{y} = W cos θ

the friction force has the formula

         fr = μ N

         fr = μ Wy

         fr = μ mg cos θ

from equation 1

at the point where the force equals the maximum friction force

in this case the block is still still so a = 0

           F = fr

           F = (μ  mg) cos θ

We can see that the quantities in parentheses with constants, so as the angle increases, the applied force must be less.

This is the force that balances the friction force, any force slightly greater than F initiates the movement.

Consequently, to overcome the out of friction we must increase the angle of the plane

the correct answer is to increase the angle of the plane

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Answer:

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Explanation:

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E_y=1175510.2\ N.C^{-1}

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Explanation:

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diagonal=\sqrt{52.5^2+52.5^2}

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E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}

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E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}

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<u>Force by each of the charges at the remaining corners:</u>

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E_y=E_1-E_4

E_y=1175510.2\ N.C^{-1}

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E_y=E_2-E_3

E_y=0\ N.C^{-1}

So the Magnitude of electric field is in the upward direction as shown directly towards the charge q_1.

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