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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Answer:
V=21.0211m/s
Explanation:
Use V=vi+at
So, V=17.46m/s+(1.49m/
)(2.39s)= 21.0211m/s
<h2>
Answer: Infrared light</h2>
A dark nebula is a cloud of dust and cold gas, which does not emit visible light and hides the stars it contains.
These types of nebulae are composed mainly of the hydrogen they obtain from nearby stars, which is their fuel.
It is using infrared light that we can "observe" and analyze in detail what happens in the inner parts of these nebulae.
To solve this problem, we will apply the concepts related to Faraday's law that describes the behavior of the emf induced in the loop. Remember that this can be expressed as the product between the number of loops and the variation of the magnetic flux per unit of time. At the same time the magnetic flux through a loop of cross sectional area is,
Here,
= Angle between areal vector and magnetic field direction.
According to Faraday's law, induced emf in the loop is,
At time 
, Induced emf is,
Therefore the magnitude of the induced emf is 10.9V
I think the answer is photosynthis, when plants turn light into food and energy.
