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masha68 [24]
3 years ago
6

A 50 mL graduated cylinder contains 25.0 mL of water. A 42.5040 g piece of gold is placed in the graduated cylinder and the wate

r level rises to 27.2 mL. What is the density of the piece of gold?
Physics
1 answer:
Elenna [48]3 years ago
5 0

Answer:

19320 kg/m³

Explanation:

density: This can be defined as the ratio of the mass of a body to its volume. The S.I unit of Density is kg/m³.

The formula of density is given as,

D = m/v ......................... Equation 1.

Where D = Density of the gold, m = mass of the gold, v = volume of the gold.

Note: From Archimedes's Principle, the piece of gold displace an amount of water that is equal to it's volume.

Amount of water displace = 27.2 - 25 = 2.2 mL.

Given: m = 42.504 g = 0.042504 kg, v = 2.2 mL = (2.2/10⁶) m³ = 0.0000022 m³

Substitute into equation 1

D = 0.042504/0.0000022

D = 19320 kg/m³

Hence the density of the piece of gold = 19320 kg/m³

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Answer:280.216j/kg°C

Explanation:

Mass of metal=0.0663kg

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Final temperature=27.4°C

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specific heat capacity of water=4186j/kg°C

0.0663xax(241-27.4)=0.395x4186x(27.4-25)

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14.16168a=3968.328

a=3968.328 ➗ 14.16168

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3 years ago
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k0ka [10]
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3 years ago
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solmaris [256]

Answer:

90 ohms

Explanation:

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take the reciprocal of 2/180 which is 180/2 and its 90 ohms

3 0
3 years ago
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Illusion [34]

Answer:

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4 0
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