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:
Regulus
Explanation:
Which is 77.63 light years away
4. hyperdermis is not a layer of skin
Answer:
velocity = 472 m/s
velocity = 52.4 m/s
Explanation:
given data
steady rate = 0.750 m³/s
diameter = 4.50 cm
solution
we use here flow rate formula that is
flow rate = Area × velocity .............1
0.750 = 
× (4.50×
)² × velocity
solve it we get
velocity = 472 m/s
and
when it 3 time diameter
put valuer in equation 1
0.750 = × 3 × (4.50×)² × velocity
velocity = 52.4 m/s
Answer:
13.7m
Explanation:
Since there's no external force acting on the astronaut or the satellite, the momentum must be conserved before and after the push. Since both are at rest before, momentum is 0.
After the push
Where 
is the mass of the astronaut, 
is the mass of the satellite, 
is the speed of the satellite. We can calculate the speed 
of the astronaut:
So the astronaut has a opposite direction with the satellite motion, which is further away from the shuttle. Since it takes 7.5 s for the astronaut to make contact with the shuttle, the distance would be
d = vt = 1.83 * 7.5 = 13.7 m
