Assignment 3 Solution - freestudy.co.uk · Assignment 3 Solution H 0.75:= B 0.2:= ρ := 1010 h1...
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Assignment 3 Solution hcyl 0.75 := h1 0.5 := ρ 1010 := B 0.2 := H 0.75 := A h1 B ⋅ := A 100 10 3 − × = D 0.3 := a π D 2 ⋅ 4 := a 70.686 10 3 − × = R ρ 9.81 ⋅ A ⋅ h1 2 ⋅ := R 247.703 10 0 × = hbar 2 h1 ⋅ 3 := hbar 333.333 10 3 − × = x H h1 − hbar + ( ) := x 583.333 10 3 − × = F Rx () ⋅ H := F 192.657 10 0 × = h2 F ρ 9.81 ⋅ a ⋅ := h2 275.083 10 3 − × = Mcyl ρ a ⋅ h2 ⋅ := Mcyl 19.639 10 0 × = Wcyl 9.81 Mcyl ⋅ := Wcyl 192.657 10 0 × = ρ cyl Mcyl a hcyl ⋅ := ρ cyl 370.445 10 0 × =
Transcript of Assignment 3 Solution - freestudy.co.uk · Assignment 3 Solution H 0.75:= B 0.2:= ρ := 1010 h1...
Assignment 3 Solution
hcyl 0.75:= h1 0.5:= ρ 1010:= B 0.2:= H 0.75:=
A h1 B⋅:= A 100 10 3−×=
D 0.3:= aπ D2⋅
4:= a 70.686 10 3−
×=
R ρ 9.81⋅ A⋅h12
⋅:= R 247.703 100×=
hbar2 h1⋅
3:= hbar 333.333 10 3−
×=
x H h1− hbar+( ):= x 583.333 10 3−×=
FR x( )⋅
H:= F 192.657 100
×=
h2F
ρ 9.81⋅ a⋅:= h2 275.083 10 3−
×=
Mcyl ρ a⋅ h2⋅:= Mcyl 19.639 100×=
Wcyl 9.81 Mcyl⋅:= Wcyl 192.657 100×=
ρcylMcyla hcyl⋅
:= ρcyl 370.445 100×=
