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zalisa [80]
3 years ago
5

A body has masses of 0.013kg and 0.012kg in oil and water respectively, if the relative density of oil is 0.875, calculate the m

ass of the body
Physics
1 answer:
konstantin123 [22]3 years ago
6 0

Answer:

the mass of the body is 0.02 kg.

Explanation:

Given;

relative density of the oil, \gamma _0 = 0.875

mass of the object in oil, M_o = 0.013 kg

mass of the object in water, M_w = 0.012 kg

let the mass of the object in air = M_a

weight of the oil, W_0 = M_a - 0.013

weight of the water, W_w = M_a - 0.012

The relative density of the oil is given as;

\gamma_0 = \frac{density \ of \ oil }{density \ of \ water} = \frac{W_0}{W_w} = \frac{M_a -0.013}{M_a -0.012} \\\\0.875 = \frac{M_a -0.013}{M_a -0.012}\\\\0.875(M_a - 0.012) = M_a - 0.013\\\\0.875M_a - 0.0105 = M_a -0.013\\\\0.875M_a - M_a = 0.0105 - 0.013\\\\-0.125 M_a = -0.0025\\\\M_a = \frac{0.0025}{0.125} \\\\M_a = 0.02 \ kg

Therefore, the mass of the body is 0.02 kg.

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An office window has dimensions 3.1 m by 2.1 m. As a result of the passage of a storm, the outside air pressure drops to 0.954 a
Virty [35]

Answer:

Net forces which pushes the window is 30342.78 N.

Explanation:

Given:

Dimension of the office window.

Length of the window = 3.1 m

Width of the window = 2.1 m

Area of the window = (3.1\times 2.1) = 6.51\ m^2

Difference in air pressure = Inside pressure - Outside pressure

                                           = (1.0-0.954) atm = 0.046 atm

Conversion of the pressure in its SI unit.

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We know,

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⇒ Pressure=\frac{Force }{Area}

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⇒ Force =4660.95\times 6.51

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The net forces which pushes the window is 30342.78 N.

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