¿Cuál es la densidad de un metal si una muestra tiene una masa de 63.5 g cuando se mide en el aire y una masa aparente de 55.4 g cuando está sumergida en agua?. Considere la densidad del agua como 1000 kg/m3.

Answers

Answer 1

La densidad del metal como se requiere en la pregunta es 7.8 * 10 ^ 3 Kg / m ^ 3.

Sabemos que ese empuje hacia arriba = peso en aire - peso en líquido

Peso en el aire = 63,5 * 10 ^ -3 Kg * 10 m / s ^ 2 = 0,635 N

Peso en líquido = 55,4 * 10 ^ -3 Kg * 10m / s ^ 2 = 0,554 N

Empuje hacia arriba = 0,635 N - 0,554 N = 0,081 N

Pero ;

Empuje hacia arriba = Volumen * densidad del fluido * aceleración debido a la gratitud

volumen = empuje hacia arriba / densidad del fluido * aceleración debido a la gratitud

volumen = 0.081 N / 1000 * 10

Volumen = 8.1 * 10 ^ -6 m ^ 3

Densidad = masa / volumen

Densidad = 63,5 * 10 ^ -3 Kg / 8,1 * 10 ^ -6 m ^ 3

= 7,8 * 10 ^ 3 kg / m ^ 3

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The equations that can be used to solve for the force of friction exerted on the block by the surface is F- f = mv² / d

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HOW TO CALCULATE SPECIFIC HEAT CAPACITY:

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Explanation:

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