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kobusy [5.1K]
3 years ago
11

If an object is not moving, what do we know about the forces acting on it?

Physics
2 answers:
Nat2105 [25]3 years ago
7 0
I think it should be 'They are balanced' if I'm not wrong
erik [133]3 years ago
7 0
If the object is not moving then the forces acting on it are balanced. Also if the object is moving in a straight line at a constant speed than the forces on it are also balanced.
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Answer:

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Explanation:

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Help on this physics question
fredd [130]

By using the equations for <em>parabolic</em> motion, we proceed to present the answers for the paragraph seen in the picture: a) t ≈ 0.553 s, b) s = 2.212 m, c) s = 11.060 m.

<h3>How to analyze a system on parabolic motion</h3>

A system is on <em>parabolic</em> motion if such system can be represented as a particle, that is, a system whose geometry is negligible, and its motion is a combination of <em>horizontal</em> movement at <em>constant</em> velocity and <em>vertical</em> <em>uniformly accelerated</em> movement due to gravity and all <em>viscous</em> and <em>rotational</em> effects are negligible.

The time required for the droplet to reach the ground is:

1.5 m = (1 / 2) · (9.807 m / s²) · t²

t = √[2 · (1.5 m) / (9.807 m / s²)]

t ≈ 0.553 s

And the <em>horizontal</em> distance traveled by the droplet is:

s = (4 m / s) · (0.553 s)

s = 2.212 m

Now, we apply the same procedure for the case of sneezing person:

1.5 m = (1 / 2) · (9.807 m / s²) · t²

t = √[2 · (1.5 m) / (9.807 m / s²)]

t ≈ 0.553 s

s = (20 m / s) · (0.553 s)

s = 11.060 m

To learn more on parabolic motion: brainly.com/question/16992646

#SPJ1

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Homocysteine is bad for the heart because it can .. ?
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B damage the protective layers of the artery walls, which can intimately lead to heart disease.
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In a double slit experiment, the intensity of light at the center of the central bright fringe is measured to be 6.2 µW/m2. What
-BARSIC- [3]

Answer:

   I_FWHW = 3.2 μW / m²

Explanation:

In the analysis of optics and electricity a very useful magnitude is the width at half height (FWHW) and the intensity at this height, which is given by

               I_FWHW = I₀ / 2

corresponds to the width of the line for this intensity.

In this case they give the maximum intensity for which

               I_FWHW = 6.2 / 2

               I_FWHW = 3.2 μW / m²

You do not give more data in your exercise, but the most interesting calculation is to find the angle values ​​for which you have this intensity since it is this range is 50% of the energy of the system, have I write the equation for this calculation

             I = Io cos² x₁   (sin x / x)²

             x₁ = π d sin θ /λ

             x = π b sin θ /λ

where d is the separation of the slits and b the width of each slit

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Which of the following represents an image that is located in front of a lens?
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-di represents an image in front of a lens
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