3 years ago

For an extended object, the weight force can be considered to act at which point?

Physics

1 answer:

JulsSmile [24]3 years ago

5 0

Answer:

At the center of the object

At the end of the object farthest away from the ground

At the center of gravity of the object

At end of the object closest to the ground

Explanation:

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Answer:

7.5 m/s²

Explanation:

Given:

v₀ = 0 m/s

v = 30 m/s

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Explanation:

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4 years ago

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A force is applied to a block sliding along a surface (Figure 2). The magnitude of the force is 15 N, and the horizontal compone

Lemur [1.5K]

Answer:

Fy = 14.3 [N]

Explanation:

To be able to solve this problem we must know that the force is a vector and has magnitude and direction, therefore it can be decomposed into the force in the X & y components:

When we have the components on the horizontal and vertical axes we must use the Pythagorean theorem.

$F = \sqrt{F_{x}^{2} +F_{y}^{2} }$

where:

F = 15 [N]

Fx = horizontal component = 4.5 [N]

Fy = vertical component [N]

$15=\sqrt{4.5^{2}+F_{y}^{2}}\\ 15^{2}= (\sqrt{4.5^{2}+F_{y}^{2}})^{2} \\225 = 4.5^{2}+F_{y} ^{2}\\ F_{y}^{2} =225 -4.5^{2}\\ F_{y}^{2}=204.75\\F_{y}=\sqrt{204.75}\\ F_{y}=14.3 [N]$

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3 years ago