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
given,
wavelength = λ = 18.7 cm
= 0.187 m
amplitude , A = 2.34 cm
v = 0.38 m/s
A) angular frequency = ?
angular frequency ,
ω = 2π f
ω = 2π x 2.03
ω = 12.75 rad/s
B) the wave number ,
C)
as the wave is propagating in -x direction, the sign is positive between x and t
y ( x ,t) = A sin(k x - ω t)
y ( x ,t) = 2.34 x sin(33.59 x - 12.75 t)
Answer:
True
Explanation:
This is because to know how someone behaves, they have to perform a particular action.
