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Sagot :
The thickness of the foil is 0.00516 mm.
To calculate the thickness of the foil using the formula of volume of a cuboid.
Formula:
- V = LWn................ Equation 1
Where:
- V = Volume of the foil
- L = Length of the foil
- W = Width of the foil
- n = Thickness of the foil.
Make n the subject of the equation
- n = V/LW.............. Equation 2
From the question,
Given:
- V = 0.645 mm³
- L = 10.0 mm
- W = 12.5 mm
Substitute these values into equation 2
- n = (0.645)/(10×12.5)
- n = 0.645/125
- n = 0.00516 mm
Hence, The thickness of the foil is 0.00516 mm.
Learn more about volume here: https://brainly.com/question/1972490
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