The lifting capacity of the elevator is 3 tons. How many sheets of iron can be loaded into the elevator if each sheet is 3 m long, 60 cm wide and 4 mm thick?
l = 3 m.
a = 60 cm = 0.6 m.
d = 4 mm = 4 * 10 ^ -3 m.
ρ = 7800 kg / m ^ 3.
Let’s find the mass of one sheet m1 by the formula: m1 = ρ * V1, where ρ is the density of iron, V1 is the volume of one sheet of iron.
Since the sheet has the shape of a rectangular parallelepiped, its volume V1 will be expressed by the formula: V1 = l * a * d, where l is the length of the sheet, a is the width of the sheet, d is the thickness of the sheet.
Then the formula for the mass of one sheet m1 will be: m1 = ρ * l * a * d.
N = m / m1 = m / ρ * l * a * d.
N = 3000 kg / 7800 kg / m ^ 3 * 3 m * 0.6 m * 4 * 10 ^ -3 m = 53.
Answer: The elevator will be able to lift N = 53 iron sheets at a time.
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