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bagirrra123 [75]
3 years ago
8

Aristotle said that a moving earthly or `mundane' object with nothing pushing or pulling on it will always A slow down and stop.

B speed up. C keep moving at the same speed. D follow a circular path.
Physics
1 answer:
andrew11 [14]3 years ago
5 0

Answer:

A slow down and stop

Explanation:

When there is no force acting on something it automatically begins to slow down and then stops.Essentially, Aristotle's perspective of motion is that "it requires a force to move an object in an unnatural" way— or, plainly, that "movement involves strength." Indeed, if you propel a book, it keeps moving. Once you stop trying to push, it comes to a stop.

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This problem must be solved using a sketch. I attached an illustration of the problem.
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Law of reflection tells us that light ray reflects off the surface at the same angle in which it falls on it( i attached another illustration of this).
Now we can write tangens equations:
tan(\theta)=\frac{h-0.8}{1}\\
tan(\theta)=\frac{1.8-h}{2.8}\\
\frac{h-0.8}{1}=\frac{1.8-h}{2.8}
We solve for h:
\frac{h-0.8}{1}=\frac{1.8-h}{2.8}\\
2.8h-0.8\cdot 2.8=1.8-h\\
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3 years ago
A _______ is a repeating disturbance or vibration that transfers or moves energy from place to place without transporting mass.
cupoosta [38]

the answer is a wave


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3 years ago
Read 2 more answers
A 26-g steel-jacketed bullet is fired with a velocity of 630 m/s toward a steel plate and ricochets along path CD with a velocit
Alecsey [184]

Answer:

  F = - 3.53 10⁵ N

Explanation:

This problem must be solved using the relationship between momentum and the amount of movement.

          I = F t = Δp

To find the time we use that the average speed in the contact is constant (v = 600m / s), let's use the uniform movement ratio

        v = d / t

        t = d / v

Reduce SI system

          m = 26 g ( 1 kg/1000g) = 26 10⁻³ kg

          d = 50 mm ( 1m/ 1000 mm) = 50 10⁻³ m

Let's calculate

         t = 50 10⁻³ / 600

         t = 8.33 10⁻⁵ s

With this value we use the momentum and momentum relationship

        F t = m v - m v₀

As the bullet bounces the speed sign after the crash is negative

       F = m (v-vo) / t

       F = 26 10⁻³ (-500 - 630) / 8.33 10⁻⁵

       F = - 3.53 10⁵ N

The negative sign indicates that the force is exerted against the bullet

5 0
3 years ago
Section 1: Experimental Overview
Luden [163]

Answer:

1-state what the lab is about, that is, what scientific concept (theory, principle, procedure, etc.) you are supposed to be learning about by doing the lab. You should do this briefly, in a sentence or two. If you are having trouble writing the opening sentence of the report, you can try something like: "This laboratory experiment focuses on X…"; "This lab is designed to help students learn about, observe, or investigate, X…." Or begin with a definition of the scientific concept: "X is a theory that…."

2-give the necessary background for the scientific concept by telling what you know about it (the main references you can use are the lab manual, the textbook, lecture notes, and other sources recommended by the lab manual or lab instructor; in more advanced labs you may also be expected to cite the findings of previous scientific studies related to the lab). In relatively simple labs you can do this in a paragraph following the initial statement of the learning context. But in more complex labs, the background may require more paragraphs.

Explanation:

7 0
3 years ago
A pressure that will support a column of Hg to a height of 256 mm would support a column of water to what height? The density of
Paul [167]

Answer:

<em>The height of water in the column = 348.14 cm</em>

Explanation:

<em>Pressure:</em><em>This is defined as the ratio of the force acting normally ( perpendicular) to the area of surface in contact. The S.I unit of  pressure is N/m²</em>

<em>p = Dgh............... Equation 1</em>

<em>Where p = pressure, D = density, g = acceleration due to gravity, h = height.</em>

<em>From the question, the same pressure will support the column of mercury and water.</em>

<em>p₁ = p₂</em>

<em>Where p₁ = pressure of mercury, p₂ = pressure of water</em>

D₁gh₁ = D₂gh₂.................. Equation 2

making h₂ the subject of equation 2

h₂ = D₁gh/D₂g............... Equation 3

Where D₁ and D₂ = Density of mercury and water respectively, h₁ and h₂ = height of mercury and water respectively

Given: D₁ = 13.6 g/cm³, D₂ = 1.00 g/cm³, h₁ = 256 mm = 25.6 cm.

Constant: g = 9.8 m/s²

Substituting these values into Equation 3,

h₂ = (13.6×9.8×25.6)/1×9.8

<em>h₂ = 348.14 cm</em>

<em>The height of water in the column = 348.14 cm</em>

6 0
3 years ago
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