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ziro4ka [17]
3 years ago
9

A sample of gold has a mass of 38.6 grams and a volume of 2 cm3 what is the density of gold

Physics
2 answers:
Contact [7]3 years ago
7 0

Answer:

Density, d=19.3\ g/cm^3

Explanation:

Mass of the sample of gold, m = 38.6 grams

Volume of gold sample, V=2\ cm^3

The density of gold sample is given by mass per unit volume. Mathematically, it is given by :

d=\dfrac{m}{V}

d=\dfrac{38.6\ g}{2\ cm^3}

d=19.3\ g/cm^3

So, the density of the sample of gold is 19.3\ g/cm^3. Hence, this is the required solution.

Rudik [331]3 years ago
3 0
Density = Mass ÷ Volume

Your mass = 38.6 grams
Your volume = 2 cm3

38.6 ÷ 2 = 19.3 grams per cm3
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The force on the segment of the wire is calculated as follows;

F = ILBsin(\theta)

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The force on the wire segment will be perpendicular in negative z-direction (applying right hand rule), so there won't be any x and y component of the force.

The angle between the wire and the magnetic field is calculated as follows;

\theta = tan^{-1} (\frac{y}{x} )\\\\\theta = tan^{-1} (\frac{4.3}{3.2} )\\\\\theta = 53.3 \ ^0

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|l | = \sqrt{3.2^2 + 4.3^2} = 5.36 \ cm = 0.0536 \ m

The component of the force in negative z-direction is calculated as;

F_z = -ILB sin(\theta)\\\\F_z = -2.7 \times 0.0536 \times 1.24 \times  sin(53.3)\\\\F_z = -0.144 \ N

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