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Pani-rosa [81]
3 years ago
11

Un cuerpo se encuentra en reposo sobre una mesa horizontal. Entonces se puede afirmar que:

Physics
1 answer:
vazorg [7]3 years ago
5 0

Answer:

C) solo III

Explanation:

Para solucionar este problema debemos analizar cada una de las opciones hasta llegar a la opcion valida.

I) el cuerpo pesa igual que su masa.

Esta opcion no puede ser ya que el peso de un cuerpo se define como el producto de la masa por la aceleracion gravitacion.

w=m*g

donde:

w = peso [N]

m = masa [kg]

g = aceleracion gravitacional = 9.81 [m/s²]

Como podemos ver el peso siempre sera mayar que la masa, ya que el peso es resultado de la multiplicacion de la masa por la gravedad.

II) Por medio de un analisis de fuerzas en el eje-y, la fuerza del peso se dirige hacia abajo mientras que la fuerza normal tiene igual magnitud, pero se dirige hacia arriba. Por esto la segunda opcion no puede ser.

III) El cuerpo se encuentra en equilibrio, es decir las unicas fuerzas que actuan sobre el cuerpo son el peso y la fuerza normal. Pero estas fuerzas son iguales y opuestas en direccion, por la tanto se cancelan y estan en equilibrio.

Esta es la opcion valida, la fuerza neta es nula.

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photoshop1234 [79]
Scientific form = 6.5 x 109.
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3 years ago
A skateboarder has an acceleration of −1.9 m/s2. If her initial speed is 6 m/s, how long does it take her to stop?
SSSSS [86.1K]
You said that she's losing 1.9 m/s of her speed every second.

So it'll take

             (6 m/s) / (1.9 m/s²)  =  3.158 seconds  (rounded)

to lose all of her initial speed, and stop.
6 0
3 years ago
When a piano tuner strikes both the A on the piano and a 440 Hz tuning fork, he hears a beat every 2 seconds. The frequency of t
AysviL [449]

Answer:

The  frequency is  f = 439.5 \  Hz        

Explanation:

From the question we are told that

   The frequency of the  tuning fork is  f_t  =  440 \ Hz

   The beat period is  T  =  2 \  s

Generally the beat frequency is mathematically represented as

       f_b  =  \frac{1}{T}

       f_b  =  \frac{1}{2}

      f_b  = 0.5 \  Hz

The  beat frequency is also represented mathematically as

     f_b  =  f_t  -  f

Where  f is the frequency of the piano

 So

       f =  440 -  0.5  

       f = 439.5 \  Hz        

3 0
3 years ago
When light propagates through two adjacent materials that have different optical properties, some interesting phenomena occur at
Veseljchak [2.6K]

Answer:

First, the different indices of refraction must be taken into account (in different media): for example, the refractive index of light in a vacuum is 1 (since vacuum = c).  The value of the refractive index of the medium is a measure of its "optical density":  Light spreads at maximum speed in a vacuum but slower in others  transparent media; therefore in all of them n> 1. Examples of typical values ​​of  are those of air (1,0003), water (1.33), glass (1.46 - 1.66) or diamond (2.42).

The refractive index has a maximum value and a minimum value, which we can calculate the minimum value by means of the following explanation:

The limit or minimum angle, α lim, is defined as the angle of refraction from which  the refracted ray disappears and all the light is reflected. As in the maximum value of  angle of refraction, from which everything is reflected, is βmax = 90º, we can  know the limit angle (the minimum angle that we would have to have to know the minimum index of refraction) by Snell's law:

 βmax = 90º ⇒ n 1x sin α (lim) = n 2 ⇒ sin α lim = n 2 / n 1

Explanation:

When a light ray strikes the separation surface between two media  different, the incident beam is divided into three: the most intense penetrates the second  half forming the refracted ray, another is reflected on the surface and the third is  breaks down into numerous weak beams emerging from the point of incidence in  all directions, forming a set of stray light beams.

4 0
3 years ago
With detailed explaniation
belka [17]
  • Ø=37°
  • Initial velocity=u=20m/s
  • g=10m/s²

#A

\\ \rm\Rrightarrow H_{max}=\dfrac{u^2sin^2\theta}{2g}

\\ \rm\Rrightarrow H_{max}=\dfrac{20^2(sin37)^2}{2(10)}

\\ \rm\Rrightarrow H_{max}=\dfrac{400sin^237}{20}

\\ \rm\Rrightarrow H_{max}=20sin^237

\\ \rm\Rrightarrow H_{max}=7.2m

#B

\\ \rm\Rrightarrow R=\dfrac{u^2sin2\theta}{g}

\\ \rm\Rrightarrow R=\dfrac{20^2sin74}{10}

\\ \rm\Rrightarrow R=40sin74

\\ \rm\Rrightarrow R=38.5m

#C

\\ \rm\Rrightarrow T=\dfrac{2usin\theta}{g}

\\ \rm\Rrightarrow T=\dfrac{2(20)sin37}{10}

\\ \rm\Rrightarrow T=4sin37

\\ \rm\Rrightarrow T=2.4s

Now

\\ \rm\Rrightarrow v=u-gt

\\ \rm\Rrightarrow v=20-10(2.4)

\\ \rm\Rrightarrow v=20-24

\\ \rm\Rrightarrow v=-4m/s

4 0
2 years ago
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