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Slav-nsk [51]
3 years ago
13

Forces of attraction between particles is the smallest in

Physics
1 answer:
olya-2409 [2.1K]3 years ago
4 0

Answer:

The forces of attraction are weak in gases.

Explanation:

Solid-state objects are presented as bodies in a definite form; their atoms are often intertwined into narrowly defined structures, which gives them the ability to withstand forces without apparent deformation. They are generally described as hard as well as resistant, and in them, the forces of attraction are greater than those of repulsion. In the crystalline solids, the presence of small intermolecular spaces gives way to the intervention of the bond forces, which place the cells in geometric forms.

Solid – In a solid, the attractive forces keep the particles together tightly enough so that the particles do not move past each other.   Their vibration is related to their kinetic energy.  In the solid the particles vibrate in place.

Liquid – In a liquid, particles will flow or glide over one another, but stay toward the bottom of the container. The attractive forces between particles are strong enough to hold a specific volume but not strong enough to keep the molecules sliding over each other.

Gas – In a gas, particles are in continual straight-line motion. The kinetic energy of the molecule is greater than the attractive force between them, thus they are much farther apart and move freely of each other.  In most cases, there are essentially no attractive forces between particles.  This means that a gas has nothing to hold a specific shape or volume.

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Which planet formed near the sun where the solar system temperatures were very hugh
Gnom [1K]

mercury is the closest planet from the sun

8 0
3 years ago
Read 2 more answers
The Cosmoclock 21 Ferris wheel in Yokohama City, Japan, has a diameter of 100 m. Its name comes from its 60 arms, each of which
emmainna [20.7K]

Answer:

Explanation:

A ) angular velocity ω = 2π / T

= 2 x 3.14 / 60

= .10467 rad / s

linear velocity v = ω R

=  .10467 x 50

= 5.23 m / s

centripetal force = m v² / R

= mg v² / gR

= 834 x 5.23² / 9.8 x 50

= 46.55 N

B )

apparent weight

= mg - centripetal force

= 834 - 46.55

= 787.45 N

C ) apparent weight

= mg + centripetal force

= 834 + 46.55

= 880.55 N.

D )

For apparent weight to be zero

centripetal force = mg

mg = mv² / R

v² = gR

= 9.8 x 50

= 490

v = 22.13 m /s

time period of revolution

= 2π R /v

2 x 3.14 x 50 / 22.13

= 14.19 s  

8 0
3 years ago
Supply the missing force necessary to achieve equilibrium. Show your work.
Mumz [18]

<u>Analysing the Question:</u>

We know that equilibrium is the state of a body when it has equal and opposite forces being applied on it

In this case, a net downward force of 496N is being applied and a net upward force of (106 + 106 + 142 + x) N

<u>Finding the missing force:</u>

Since we have to achieve equilibrium, the net upward forces have to be equal to the net downward forces

So,  (106 + 106 + 142 + x) = 496

354 + x = 496

x = 496 - 354

x = 142 N

Therefore, the missing force is 142 N

8 0
3 years ago
El espectro visible en el aire está comprendido entre la longitud de onda 450 nm del color azul, Determina la velocidad de propa
TiliK225 [7]

Answer:

v = 2,99913 10⁸ m / s

Explanation:

The velocity of propagation of a wave is

         v = λ f

in the case of an electromagnetic wave in a vacuum the speed that speed of light

         v = c

When the wave reaches a material medium, it is transmitted through a resonant type process, whereby the molecules of the medium vibrate at the same frequency as the wave, as the speed of the wave decreases the only way that they remain the relationship is that the donut length changes in the material medium

         λ = λ₀ / n

where n is the index of refraction of the material medium.

Therefore the expression is

           v = \frac{\lambda_o}{n} f

Let's look for the frequency of blue light in a vacuum

           f =\frac{c }{\lambda_o}  

           f = \frac{3 \ 10^8}{450 \ 10^{-9}}

           f = 6.667 10¹⁴ Hz

the refractive index of air is tabulated

           n = 1,00029

let's calculate

           v = \frac{450 \ 10^{-9} }{1.00029}  \ 6.667 \ 10^{14}450 10-9 / 1,00029 6,667 1014

            v = 2,99913 10⁸ m / s

we can see that the decrease in speed is very small

8 0
3 years ago
A 220 g mass is on a frictionless horizontal surface at the end of a spring that has force constant of 7.0
Talja [164]

The concept of conservation of energy and harmonic motion allows to find the result for the power where the kinetic and potential energy are equal is:

        x = 0.135 cm

Given parameters

  • The mass m = 220 g = 0.220 kg
  • The spring cosntnate3 k = 7.0 N / m
  • Initial displacement A = 5.2 cm = 5.2 10-2 m

To find

  • The position where the kinetic and potential energy are equal

 

A simple harmonic movement is a movement where the restoring force is proportional to the displacement, the result of this movement is described by the expression.

          x = A cos wt + fi

          w² = \frac{k}{m}

Where x is the displacement from the equilibrium position, A the initial amplitude of the system, w the angular velocity t the time, fi a phase constant determined by the initial conditions, k the spring constant and m the mass.

The speed is defined by the variation of the position with respect to time.

       v = \frac{dx}{dt}

let's evaluate

       v = - A w sin (wt + Ф)

Since the body releases for a time t = 0 the velocity is zero, therefore the expression remains.

       0 = - A w sin Ф

For the equality to be correct, the sine function must be zero, this implies that the phase constant is zero

        x = A cos wt

Let's find the point where the kinetic and potential energy are equal.

        K = U

        ½ m v² = m g x

       

we substitute

        ½ A² w² sin² wt = g A cos wt

        sin² wt = \frac{2g}{A}  cos wt

let's calculate

      w = \sqrt{\frac{7}{0.220} }  

      w = 5.64 rad / s

      sin² 5.64t = 2 9.8 / 0.052 cos 5.64t

      sin² 5.64t = 376.92 cos 5.64 t

      1 - cos² 5.64t = 376.92 cos 5.64t

      cos² 5.64t -376.92 cos564t -1 = 0

we make the change of variable

       x = cos 5.64t

      x²- 376.92 x - 1 = 0

      x = 0.026

      cos 5.64t = 0.026

   

Let's find the displacement for this time

       x = 5.2 10-2 0.026

       x = 1.35 10-3 m

In conclusion Using the concepts of conservation of energy and harmonic motion we can find the result for the could where the kientic and potential enegies are equal is:

        x = 0.135 cm

Learn more here: brainly.com/question/15707891

8 0
3 years ago
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