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Imagine a rectangular prism of pure water whose density (r) is 1.0 g/cm . The base of this imaginary prism is 1 cm on a side, and the remaining side of the prism is the water depth (a) measured in cm. The mass (m) of that prism of water, expressed in grams, is related to the water depth (a) in the following way: the value of a is __________ the value of m.

Respuesta :

ajeigbeibraheem

Answer:

equal to

Step-by-step explanation:

Given that:

base of the imaginary prism is 1 cm on a side,

the remaining side = a cm

Thus, the volume of the prism is:  = (1 × 1 × a) cm

= a cm³

also density (r) of the rectangular prism = 1.0 g/cm

recall that: density = mass/volume

mass = density × volume

mass = 1.0 g/cm × a cm³

mass = (a) g.cm²

Thus, the value of depth(a) is equal to the value of mass (m).

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