¿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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Explanation:

Correct option is

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2

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Please help! Question is attached!​

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Question 1 of 11
The book has a mass of 2.5 kg.
What net force must act on the book to make it accelerate to the left at a rate
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OB. F = 2.8 N right
O C. F = 2.8 N left
O D. F = 17.5 N left

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Answer:

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To  solve a problem using the equation for Kepler's third law, Enrico must convert the average distance of Mars from the Sun from meters into astronomical units. This is done by dividing the average distance by 1. 5 × 10^11.

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