Dr. Jones performed an experiment to monitor the effects of sunlight exposure on stem density in aquatic plants. In the study, Dr. Jones measured the mass and volume of stems grown in 5 levels of sun exposure. The data is represented below. Sun exposure Stem mass (g) Stem volume (mL)
30 275 1100
45 415 1215
60 563 1425
75 815 1610
90 954 1742
a. Convert the mass measurements to kilograms (kg) and the volume measurements to cubic meters (mº).
b. Calculate the density of the samples using the equation d = m/v. d = density m = mass (kg) v = volume (m)
c. Convert the density values to scientific notation.

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

Conservation of volume

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

Where, m = mass

V = volume

If the m = 0.275 kg and V = 0.0011 m³

Put the value into the formula

If the m = 0.415 kg and V = 0.001215 m³

Put the value into the formula

If the m = 0.563 kg and V = 0.001425 m³

Put the value into the formula

If the m = 0.815 kg and V = 0.001610 m³

Put the value into the formula

If the m = 0.954 kg and V = 0.001742 m³

Put the value into the formula

(c). We need to convert the density values to scientific notation

In scientific notation

The densities are

Hence, This is required solution.

it is very well related to a lot of things but with my limited knowledge i can confidently say that the answer is space.

the answer is b

explanation: