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maria [59]
2 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]2 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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Answer:

True.

Explanation:

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2 years ago
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6.5e-4 m

Explanation:

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Hence from Energy at the bottom of the incline= energy at the point where the block will stop

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⇒PE2-PE1=\frac{1}{2} kx^{2}

Also PE2-PE2=mgh

where m is the mass of block

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h is the difference in height between two positions

⇒mgh=\frac{1}{2} kx^{2}

Given m=2100kg

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⇒h=\frac{13.31}{20580}

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