The answer is: " 5 g / cm³ " .
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Explanation:
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Density = mass divided by volume ; or: "D = m / V " ;
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and is expressed as: "mass per unit volume" ;
The mass, "m", is expressed in units of "g" (grams) ; and
the volume, "V" is expressed in units of "cm³ " or "mL" ; ("cm³ ", in this case);
{Note the exact conversion: " 1 cm³ = 1 mL " .} .
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So, if: the mass, "m = 100 g" {given} ;
and the volume, " V = 20 cm³ " {given} ;
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Plug these values into the formula/equation to solve for the density, "D" ;
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D = m / V = (100 g) / (20 cm³)
= (100 ÷ 20) g / cm³ = 5 g /cm³ .
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The answer is: " 5 g / cm³ " .
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Answer:
q = 0.0392 / V, for V= 0.1V q = 0.392 C
Explanation:
For this exercise we can assume that the power energy of the drops is transformed into kinetic energy, therefore we use the conservation of energy
starting point
Em₀ = U = q V
final point
Em_f = K = ½ m v²
Em₀ = Em_f
q V = ½ m v²
q =
let's calculate
q = ½ 0.40 10⁻³ 14² / V
q = 0.0392 / V
The object to be painted is connected to ground therefore its potential is dro, but the gun where it is painted has a given potential, suppose it is
V = 0.1 V
q = 0.0392 / 0.1
q = 0.392 C
Answer:
the mark of the broken end is 2.6 cm so, we use the scale from the next full mark i.e. 3cm
Explanation:
<em>we </em><em>now </em><em>measure</em><em> </em><em>the </em><em>length</em><em> </em><em>of </em><em>the </em><em>pencil</em><em> </em><em>by </em><em>keeping </em><em>the </em><em>3</em><em> </em><em>c</em><em>m</em><em> </em><em>mark </em><em>of </em><em>the </em><em>scale</em><em> </em><em>at </em><em>it's</em><em> </em><em>left </em><em>end.</em>
<em>The </em><em>3</em><em> </em><em>cm </em><em>value </em><em>is </em><em>then </em><em>subtracted</em><em> </em><em>from </em><em>the </em><em>scale</em><em> </em><em>reading</em><em> </em><em>at </em><em>the </em><em>right</em><em> </em><em>side </em><em>end </em><em>of </em><em>the </em><em>pencil</em><em> </em><em>to </em><em>obtain </em><em>the </em><em>correct</em><em> </em><em>length</em><em> </em><em>of </em><em>the </em><em>pencil.</em><em> </em><em>✏️</em>
<em>(</em><em>i </em><em>i </em><em>)</em><em> </em>place the scale in the contact with object along it's length
(2) Your eyes must be exactly in front of the point where the measurements to be taken.
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Answer:
The most common oxidation numbers for a given element