A droplet of pure mercury has a density of 13.6 g/cm3. What is the density of a sample of pure mercury that is 10 times as large as the droplet?
Answer: In this case the density will remain constant for both droplets. The reason being that volume will not change the density of the material. The only way of changing it is by changing its state. If you increase the volume then the mass will also increase. Leaving the density the same.
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Answer:
the potential energy a massive object has in relation to another massive object due to gravity
Answer:
326149.2 KJ
Explanation:
The heat transfer toward and object that suffered an increase in temperature can be calculated using the expression:
Q = m*cv*ΔT
Where m is the mass of the object, cv is the specific heat capacity at constant volume, which basically means the amount of heat necessary for a 1kg of water to increase 1C degree in temperatur, and ΔT is the change in temperature.
A 65000 L swimming pool will have a mass of:
65000L * = 65000 kg
The specific heat capacity at constant volume of water is equal to 4.1814 KJ/KgC.
We replace the data and get:
Q = m*cv*ΔT = 65000 kg * 4.1814 KJ/KgC * 1.2°C = 326149.2 KJ