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weeeeeb [17]
3 years ago
12

Una masa de 0,5 kg está sobre una pendiente inclinada 20º sujeta mediante una cuerda paralela a la pendiente que impide que desl

ice. Si no hay rozamiento, ¿qué fuerza hace la cuerda?
Physics
1 answer:
jeka943 years ago
4 0

Answer:

<em>4.61 N</em>

<em></em>

Explanation:

masa = 0.5 kg

ángulo de inclinación = 20°

Peso normal de la masa = mg

donde m = masa

g = aceleración debido a la gravedad = 9.81 m/s^2

Peso normal = 0.5 x 9.81 = 4.905 N

Si la masa se mantiene en su lugar mediante una cuerda paralela al plano, y no hay fricción en la masa, entonces

La fuerza sobre la cuerda = peso normal x cos ∅

donde ∅ = 20 °

La fuerza sobre la cuerda = 4.905 x cos 20°

==><em> 4.61 N</em>

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A 5.00 kg rock whose density is 4300 kg/m3 is suspended by a string such that half of the rock's volume is under water. You may
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Answer:

The tension on the string is  T  =  43.302 \ N

Explanation:

From the question we are told that

    The mass of the rock is m_r = 5.00 \ kg =  5000 \ g

       The density of the rock is \rho  =  4300 \ kg/m^3 =  4.3 g/dm^3

       

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substituting values

        V    =  \frac{5000}{4.3}

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The volume of the rock immersed in water is

      V_w = \frac{V}{2}  

substituting values

     V_w = \frac{1162.7 }{2}

     V_w = 581.4 \ dm^3

mass of water been displaced by the this volume is

     m_w  = V_w     According to Archimedes principle

=>   m_w =  581.4 \ g

     m_w =  0.5814 \ kg

The weight of the water displace is  

      W _w =  m_w  * g

      W _w =  0.5814  * 9.8

      W _w = 5.698 \ N

The actual weight of the rock is  

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     W_r  =  5.0 *  9.8

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The tension on the string is

       T  = W_r - W_w

substituting values

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Explanation:

Hope this helps.

Please mark my answer as brainliest?

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