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SomeMichGuy · M
No.
1) You calculated mass, not weight. To obtain the weight, you have to multiply by the acceleration due to gravity...if any.
2) This is not only applicable to (mild) steel plates.
3) Are you suggesting that it is a surprise that, since the mass density, ρ, is defined to be
ρ = (dm/dτ)
where dτ is a differential volume element, leads to
M = \int_{0}^{M} dm
= \int_{0}^{V} ρ dτ
for finding the total mass M in a volume V of a substance with a given ρ...?
1) You calculated mass, not weight. To obtain the weight, you have to multiply by the acceleration due to gravity...if any.
2) This is not only applicable to (mild) steel plates.
3) Are you suggesting that it is a surprise that, since the mass density, ρ, is defined to be
ρ = (dm/dτ)
where dτ is a differential volume element, leads to
M = \int_{0}^{M} dm
= \int_{0}^{V} ρ dτ
for finding the total mass M in a volume V of a substance with a given ρ...?
KiwiBird · 36-40, F
Twice Half the Weight.
Abstraction · 61-69, M
@Gusman She's right. Applies to any steel plate.
SW-User
Interesting! What is the heaviest you've calculated?
Gusman · 61-69, M
@SW-User 6 metres x 3 metres x 50 mm x 7.85 = 7065 kilograms