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8 days ago

The density of water is 1 g/cc. Ice floats on water with 90% of its volume

Physics

2 answers:

Sav [3.1K]8 days ago

8 0

Respuesta:

Explicación:

Es bastante sencillo calcularlo al darse cuenta de que la masa del agua desplazada es igual a la masa del objeto que flota en ella.

Tomemos un iceberg de volumen V. Tiene una densidad d_i. Por lo tanto, su masa total es V*d_i = m.

Ahora, está desplazando el 90% de su volumen. El agua del mar tiene una d_w. La masa total del agua desplazada es 0.9 * V * d_w, y es igual a la masa del iceberg, V*d_i.

Dividiendo ambos lados por V. Obtienes 0.9 * d_w = d_i. Ahora solo tenemos que resolver para d_w. Dividiendo ambos lados por 0.9, obtienes d_w = d_i/0.9.

Sav [3.1K]8 days ago

3 0

Respuesta: 24.3 g

Explicación:

De acuerdo con el principio de Arquímedes,

la fuerza de flotación = peso del líquido desplazado

La densidad del objeto que flota en un líquido = densidad del líquido * fracción del objeto dentro del líquido

Matemáticamente representado así:

ρ(objeto) = ρ(agua) * fracción f

ρ(agua) = 1 g/cc, fracción del objeto dentro del líquido = 10% = 0.1

ρ(objeto) = 1 × ( 1 — 0.1 ) = 0.9 g/cc

ρ(objeto) = 0.9 g/cc

Para calcular la masa de un cubo de hielo, utilizamos la fórmula M = ρV

ρ = 0.9 g/cc, L = 3cm, V = L^3 = 3^3 = 27 cc

M = 0.9 * 27

M = 24.3 g

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On its own, a certain tow-truck has a maximum acceleration of 3.0 m/s2. what would be the maximum acceleration when this truck w

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Since the engine remains unchanged, the force generated stays the same, allowing us to set the force equations equal to each other:
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1 month ago

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Answer:

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σ₂ = $7.6 * 10 ^{-6}$ C/m²

Explanation:

Provided Information:

i) Smaller sphere's radius ( r ) = 5 cm.

ii) Larger sphere's radius ( R ) = 12 cm.

iii) Electric field at the larger sphere's surface ( E₁ ) = 358 kV/m, which is equivalent to 358 * 1000 v/m

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V = constant

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