Respuesta :
Answer:
[tex]0.05\times 10^{-3}\ \text{m}^3[/tex]
[tex]800\ \text{kg/m}^3[/tex]
Explanation:
[tex]W_b[/tex] = Weight of bottle = 0.25 N
[tex]W_w[/tex] = Weight of water = [tex]0.75-0.25=0.5\ \text{N}[/tex]
g = Acceleration due to gravity = [tex]10\ \text{m/s}^2[/tex]
[tex]\rho_w[/tex] = Density of water = [tex]1000\ \text{kg/m}^3[/tex]
Mass of water
[tex]m_w=\dfrac{W_w}{g}\\\Rightarrow m_w=\dfrac{0.5}{10}\\\Rightarrow m_w=0.05\ \text{kg}[/tex]
Volume is given by
[tex]V_w=\dfrac{m_w}{\rho_w}\\\Rightarrow V_w=\dfrac{0.05}{1000}\\\Rightarrow V_w=0.05\times 10^{-3}\ \text{m}^3[/tex]
Volume of the bottle is [tex]0.05\times 10^{-3}\ \text{m}^3[/tex].
This means that the alcohol will take the same amount of volume.
[tex]W_a[/tex] = Weight of alcohol = [tex]0.65-0.25=0.4\ \text{N}[/tex]
Mass of alcohol
[tex]m_a=\dfrac{0.4}{10}\\\Rightarrow m_a=0.04\ \text{kg}[/tex]
Density is given by
[tex]\rho=\dfrac{m_a}{V_w}\\\Rightarrow \rho=\dfrac{0.04}{0.05\times 10^{-3}}\\\Rightarrow \rho=800\ \text{kg/m}^3[/tex]
Density of alcohol is [tex]800\ \text{kg/m}^3[/tex]
The volume of the bottle is 50 mL and the density of alcohol is 0.8 g/c[tex]m^{3}[/tex]
The weight of empty bottle is 0.25N
the weight of bottle filled with water is 0.75N
therefore, the weight of water in the bottle is 0.75N - 0.25N = 0.50N
we know that weight, w =mg , where m is the mass of substance and g is acceleration due to gravity. Here we will take g= 10 m/[tex]s^{2}[/tex] for simplicity.
⇒ m = w/g = 0.5/10 = 0.05 kg
⇒ mass of water is 50g
now, the density of water ρ(w) = 1g/[tex]cm^{3}[/tex] = 1g/mL
relation between mass, density and volume is:
V = m/ρ = 50/1 = 50mL
Hence the volume of the bottle is 50mL.
Now it is given that the weight of bottle filled with alcohol is 0.65N
⇒ weight of alcohol = 0.40N
⇒ mass of alcohol = 0.40/g = 0.40/10 = 0.040kg = 40g
The volume of the bottle remains the same, so to calculate the density of alcohol we use the above mentioned relation between mass, density, and volume on alcohol.
That is, V = m/ρ
⇒ ρ(al) = m/V = 40/50 g/c[tex]m^{3}[/tex]
ρ(al) = 0.8 g/c[tex]m^{3}[/tex] or 0.8 g/mL
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