Respuesta :
Given that,
Sun exposure = 30%, 45%, 60%, 75%, 90%
Stem mass (g) = 275, 415, 563, 815, 954
Stem volume (ml) = 1100, 1215, 1425, 1610, 1742
(a). We need to convert the mass measurements to kilograms (kg) and the volume measurements to cubic meters
Using conversion of mass
[tex]1\ g=0.001\ kg[/tex]
Conservation of volume
[tex]1\ Lt=0.001\ m^3[/tex]
[tex]1\ mL=1\times10^{-6}\ m^3[/tex]
So, mass in kg
Stem mass (kg) = 0.275, 0.415, 0.563, 0.815, 0.954
Volume in m³,
Stem volume (m³) = 0.0011, 0.001215, 0.001425, 0.001610, 0.001742
(b). We need to calculate the density of the samples
Using formula of density
[tex]\rho=\dfrac{m}{V}[/tex]
Where, m = mass
V = volume
If the m = 0.275 kg and V = 0.0011 m³
Put the value into the formula
[tex]\rho=\dfrac{0.275}{0.0011}[/tex]
[tex]\rho=250\ kg/m^3[/tex]
If the m = 0.415 kg and V = 0.001215 m³
Put the value into the formula
[tex]\rho=\dfrac{0.415}{0.001215}[/tex]
[tex]\rho=341.56\ kg/m^3[/tex]
[tex]\rho=342\ kg/m^3[/tex]
If the m = 0.563 kg and V = 0.001425 m³
Put the value into the formula
[tex]\rho=\dfrac{0.563}{0.001425}[/tex]
[tex]\rho=395.08\ kg/m^3[/tex]
If the m = 0.815 kg and V = 0.001610 m³
Put the value into the formula
[tex]\rho=\dfrac{0.815}{0.001610}[/tex]
[tex]\rho=506.21\ kg/m^3[/tex]
If the m = 0.954 kg and V = 0.001742 m³
Put the value into the formula
[tex]\rho=\dfrac{0.954}{0.001742}[/tex]
[tex]\rho=547.6\ kg/m^3[/tex]
[tex]\rho=548\ kg/m^3[/tex]
(c). We need to convert the density values to scientific notation
In scientific notation
The densities are
[tex]\rho\ (kg/m^3)= 2.50\times10^{2}, 3.42\times10^{2}, 3.95\times10^{2}, 5.06\times10^{2}, 5.48\times10^{2}[/tex]
Hence, This is required solution.
