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vladimir2022 [97]
1 year ago
15

A solid weighs 237.5 g in air and 12.5 g in a liquid in which it is wholly submerged. The density of the liquid is 900 kg/m³.

Physics
1 answer:
djyliett [7]1 year ago
4 0

(i) The density of the solid is 950 kg/m³

(ii) the density of another liquid in which the same solid would float is 200 kg/m³

The given parameters;

mass of the solid in air, Ma = 237.5 g

mass of the solid in liquid, Ms = 12.5 g

density of the liquid, \rho_l = 900 \ kg/m^3

(i) The density of the solid is calculated as follows;

Weight in air - Weight in liquid = upthrust

\frac{\rho_o}{\rho_l}  = \frac{Ma}{M_a - Ms} \\\\\frac{\rho_o}{900}  = \frac{237.5}{237.5 - 12.5}\\\\\frac{\rho_o}{900}  =1.0556\\\\\rho_o = 1.0556 \times 900\\\\\rho_o = 950 \ kg/m^3

(ii) the density of another liquid in which the same solid would float with one-fifth of its volume exposed above the liquid surface.​

This is a low density liquid, because one-fifth of the solid is above the liquid surface while four-fifth is below the liquid surface.

the specific gravity of the liquid = ¹/₅ = 0.2

density of water , \rho_w = 1000 \ kg/m^3

S.G = \frac{\rho_l}{\rho_w} \\\\0.2 = \frac{\rho_l}{1000} \\\\\rho_l = 0.2 \times 1000\\\\\rho_l = 200 \ kg/m^3

Thus, the density of another liquid in which the same solid would float is 200 kg/m³

Learn more here: brainly.com/question/14400760

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Explanation:

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1 year ago
Which of the following statements about electric field lines associated with electric charges is false? Electric field lines can
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Answer:

The false statement is 'Electric field lines form closed loops'.

Explanation:

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8 0
2 years ago
The energy from 0.015 moles of octane was used to heat 250 grams of water. The temperature of the water rose from 293.0 K to 371
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Answer : The correct option is, (B) -5448 kJ/mol

Explanation :

First we have to calculate the heat required by water.

q=m\times c\times (T_2-T_1)

where,

q = heat required by water = ?

m = mass of water = 250 g

c = specific heat capacity of water = 4.18J/g.K

T_1 = initial temperature of water = 293.0 K

T_2 = final temperature of water = 371.2 K

Now put all the given values in the above formula, we get:

q=250g\times 4.18J/g.K\times (371.2-293.0)K

q=81719J

Now we have to calculate the enthalpy of combustion of octane.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy of combustion of octane = ?

q = heat released = -81719 J

n = moles of octane = 0.015 moles

Now put all the given values in the above formula, we get:

\Delta H=\frac{-81719J}{0.015mole}

\Delta H=-5447933.333J/mol=-5447.9kJ/mol\approx -5448kJ/mol

Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.

5 0
2 years ago
Ronnie kicks a playground ball with an initial velocity of 16 m/s at an angle of 40° relative to the ground. What is the approxi
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