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A sheet of aluminum measures 52.5 cm by 23.2 cm . if the volume is 2.95 cm3 what is the thickness of the foil?

User Moulde
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Answer:We can use the formula for the volume of a rectangular solid, V = lwh, where l, w, and h represent the length, width, and height of the solid, respectively. In this case, we are looking for the thickness of the aluminum sheet, so we can use h to represent the thickness.

Given:

l = 52.5 cm

w = 23.2 cm

V = 2.95 cm^3

We need to solve for h:

V = lwh

2.95 cm^3 = (52.5 cm)(23.2 cm)(h)

Divide both sides by (52.5 cm)(23.2 cm):

h = 2.95 cm^3 / (52.5 cm)(23.2 cm)

h = 0.00252 cm or 0.0252 mm (rounded to four decimal places)

Therefore, the thickness of the aluminum sheet is approximately 0.0025 cm or 0.0252 mm.

Explanation:I'm just built different

User Youi
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