When you look vertically downwards into a swimming pool full of water the bottom of the pool seems closer to you than it really is. The apparent depth at this point can be worked out using the formula:
Apparent depth = Real depth/Refractive index of water
However as you look towards the end of the pool the bottom seems to curve upwards – we are thinking about a pool of constant depth to start with.
If the depth of the pool really does change then the situation becomes more complex and you would need to allow for that when drawing the diagram
Answer:
Petroleum
Explanation:
Because Petroleum reservoirs can be depleted but something like wind energy wind will always be there
the answer is (c) i think
Answer:
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No; the sample could not be aluminum;
since the density of aluminum, " 2.7 g/cm³ " , is NOT close enough to the density of the sample, " 3.04 g/cm³ " .
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Explanation:
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Density is expressed as "mass per unit volume" ;
in which:
"mass, "m", is expressed in units of "g" (grams); and:
"Volume, "V", is expressed in units of "cm³ " (such as in this problem); or in units of "mL" ;
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{Note the exact conversion: " 1 cm³ = 1 mL " .}.
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The formula for density: D = m/V ;
Given: The density of aluminum is: 2.7 g/cm³.
Given: A sample has a mass of 52.0 g ; and Volume of 17.1 cm³ ; could it be aluminum?
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Let us divide the mass of the sample by the volume of the sample;
by using the formula:
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D = m / V ;
and see if the value is at, or very close to "2.7 g/cm³ ".
If it is, then it could be aluminum.
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The density for the sample:
D = (52.0 / 17.1) g/cm³ = 3.0409356725146199 g/cm³ ;
→round to "3 significant figures" ;
= 3.04 g/cm³ .
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No; the sample could not be aluminum; since the density of aluminum,
"2.7 g/cm³ " is NOT close enough to the density of the sample,
"3.04 g/cm³ " .
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