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maria [59]
3 years ago
7

Un alambre de teléfono de 120m de largo y de 2.2mm de diámetro se estira debido a una fuerza de 380 N cual es el esfuerzo longit

udinal si la longitud después de ser estirado es de 0.10m ¿cual es la deformación longitudinal?
Physics
1 answer:
sergij07 [2.7K]3 years ago
6 0

Respuesta: verifique amablemente la explicación

Explicación:

Dado lo siguiente:

Longitud (L) del cable = 120 m

Diámetro (d) = 2,2 mm (2,2 / 1000) = 2,2 * 10 ^ -3 m

Fuerza (F) = 380 N

Esfuerzo longitudinal = Fuerza / Área

Área = πd² / 4 = (π * (2.2 * 10 ^ -3) ^ 2) / 4

Área = (3.142 * 4.84 * 10 ^ -6)

Área = 0.00000380132 m²

Estrés = Fuerza / Área

Estrés = 380 / 0.00000380132

Esfuerzo longitudinal = 99952128.12 = 9.9952128 * 10^7 Nm^-2

Deformación longitudinal: extensión / longitud

Extensión = 0.10 m

Longitud = 120 m

Deformación longitudinal = 0,1 m / 120 m

Deformación longitudinal = 0.0008333 = 8.33 × 10 ^ -4

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A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be wh
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Answer:

The value is  v  =  2.3359 *10^{4} \ m/s

Explanation:

From the question we are told that

  The  initial speed is u =  2.05 *10^{4} \  m/s

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             T__{E}} =  KE__{i}} +  KE__{e}}

Here  KE_i is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

          KE_i =   \frac{1}{2}  *  m  *  u^2

=>       KE_i =   \frac{1}{2}  *  m  *  (2.05 *10^{4})^2

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And  KE_e is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

       KE_e =  \frac{1}{2}  *  m *  v_e^2

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