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denpristay [2]
2 years ago
10

Kayla starts at -3, walks 5 blocks right and 3 blocks left. What is her displacement?

Physics
1 answer:
Elanso [62]2 years ago
7 0

Answer: The displacement is 1 block.

Explanation:

Let's define:

The right is the positive side.

The left is the negative side.

Then if you start at position A, and you walk N blocks to the right, the new position is:

A + N

And if you start at position A, and you walk M blocks to the left, the new position is:

A - M.

In this case, we know that Kayla starts at -3 and she walks 5 blocks to the right.

Then her new position is:

-3 + 5 = 2

Now she walks 3 blocks to the left, then her new position is:

2 - 3 = -1

The displacement will be equal to the difference between the final position (-1) and the initial position (-2)

Then the displacement is:

D = -1 - (-2) = -1 +2 = 1

The displacement is 1 block.

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Reason:

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Magnitude of the adhesive = F_{adhesion}

Weight of the top block = W_{top}

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Given that with the exertion of the tension, the two blocks remain at rest, we have;

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The adhesive causes the bottom block to remain attached to the top block, we have;

Therefore, the magnitude of the adhesive force adds the bottom weight, to the top, weight, which gives;

The magnitude of the adhesive force = The weight of the bottom block

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Four forces are exerted on a disk of radius R that is free to spin about its center, as shown above. The magnitudes are proporti
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The given magnitude of forces of F₁ = F₄, F₂ = F₃, F₁ = 2·F₂, give the

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The given parameters are;

F₁ = F₄

F₂ = F₃

F₁ = 2·F₂

Therefore;

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In vector form, we have;

\vec{F_4} = \mathbf{\frac{\sqrt{3} }{2} \cdot F_4 \cdot \hat i -  0.5 \cdot F_4 \hat j}

\vec{F_2} = \mathbf{ -F_2 \,  \hat j}

Clockwise moment due to F₄, M₁ = -0.5 \times F_4 \,  \hat j  \times \dfrac{R}{2}

Therefore;

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Counterclockwise moment due to F₂ = -F_2 \,  \hat j  \times \dfrac{R}{2}

Given that the clockwise moment due to F₄ = The counterclockwise moment due to F₂, we have;

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