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aalyn [17]
4 years ago
7

Cual es la fuerza de empuje que recibe un cuerpo de forma cilíndrica que mide 0.3 m de radio y 8m de altura, el cual se sumerge

en alcohol si su densidad es de 790 Km/m3 tomando en cuenta la gravedad de 9.8m/seg2
Physics
1 answer:
zimovet [89]4 years ago
3 0

Responder:

Fb = 17,5 kN

Explicación:

La expresión de la fuerza de empuje

Fb = ρgV ---------- 1

dónde

Fb = fuerza de flotación

ρ = densidad del fluido

g = aceleración debida a la gravedad

V = volumen de fluido

Paso uno:

datos dados

radio del cilindro = 0,3 m

altura h = 8m

densidad = 790 kg / m ^ 3

g = 9,8 m / s ^ 2

El volumen del cilindro se expresa como

V = πr ^ 2h

V = 3,142 * 0,3 ^ 2 * 8

V = 3,142 * 0,09 * 8

V = 2,26

El volumen es 2.26m ^ 3, el volumen del cilindro es igual al volumen de alcohol disparado

La expresión de la fuerza de empuje es

Fb = ρgV

sustituyendo tenemos

Fb = 790 * 9,8 * 2,26

Fb = 17514,26N

Fb = 17,5 kN

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