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alexdok [17]
3 years ago
13

A merchant in Katmandu sells you a solid gold 1.00-kg statue for a very reasonable price. When you get home, you wonder whether

or not you got a bargain, so you lower the statue into a container of water and measure the volume of displaced water.
A) Find the volume of water that will be displaced for pure gold.
B) For my answer I found 51813 kg/m^3. It marked me down saying enter your answer using dimension of volume.... i thought kg/m^3 was dimesnsion of volume?
Physics
1 answer:
White raven [17]3 years ago
7 0

Answer:

a) 51.8 cm³

b) kg/m³ is a dimension of density (mass/volume). The regular unitys for volume are m³, cm³, L, gallons.

Explanation:

a) The density of pure gold is 19.3 g/cm³. When put in water, the piece of gold will occupy a volume, so that the volume of water will be displaced. To know the volume, we must divide the mass for the density (mass must be in grams because of the units of the density)

V = 1000/19.3

V = 51.8 cm³

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Suppose the battery in a clock wears out after moving thousand coulombs of charge through the clock at a rate of 0.5 Ma how long
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Answer:

Hello your question is poorly written below is the complete question

Suppose the battery in a clock wears out after moving Ten thousand coulombs of charge through the clock at a rate of 0.5 Ma how long did the clock run on does battery and how many electrons per second slowed?

answer :

a) 231.48 days

b) n = 3.125 * 10^15

Explanation:

Battery moved 10,000 coulombs

current rate = 0.5 mA

<u>A) Determine how long the clock run on the battery. use the relation below</u>

q = i * t ----- ( 1 )

q = charge , i = current , t = time

10000 = 0.5 * 10^-3 * t

hence  t = 2 * 10^7 secs

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<u>B) Determine how many electrons per second flowed </u>

q = n*e ------ ( 2 )

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n ( number of electrons ) = q / e = ( 0.5 * 10^-3 ) / ( 1.6 * 10^-19 )

hence : n = 3.125 * 10^15

8 0
3 years ago
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