Câu 1 ( o,g /cm3 ) o = m (g/cm3 ),(kg/dm3 ),(T/m3 ) o ->m1=m+mparafin .ngâm ng ->FAS = ΔP(m1−m2)g (1) FAS = (Vo +Vparafin).n.g (2) Vparafin).n..g(m1−m2).g  = m (g /cm3 ) m1−m2 − m1−m  n parafin  m1−m2 m1−m2 o+Vparafin o parafin n n n =1g /cm3 o = n m−m2 ->m1 -mô`ng - ∈  o + ∈ o = m1−mô`ng ô`ng 3 -BT: = 1,8 -2,5g/cm -  = 0,5 - 1,8 3 g/cm -  <0,5 3 g/cm -  =2,5 - 5 3 g/cm - - 25 MPa - - 50 MPa 1 -BT cao : 60 - 100 MPa - - -- - - - H2O 3CaO.SiO2 , 2Ca.SiO2,3Ca.Al2O ,4Ca.Al2O .Fe2O i: - - - -- -  -2cm.   2 -88%), h(8-11%),,O(0-1,5%),,S(0-6%),,N(0- to max - F (N/mm3 = MPa) pmax - Ri tb n Rtb - - - -- - - -3cm) 3 - 10 cm -4 cm - 20 cm - -- ->ch -0,6% -2% ->Fe  6-10.  12-42. - Δl =l1 −l0 4  = Δl.100% 0  = F0 − Fk .100% 0   Edh = đh đh  0<1 <2  - : 1 = 1 1 P=P22 = P 2 Eđh = tg = 2 −1 2 1 - -  = E. - -  -- - Bê tôn 3 -BT: = 1,8 -2,5g/cm -  = 0,5 - 1,8 3 g/cm -  <0,5 3 g/cm -  =2,5 - 5 3 g/cm - - 25 MPa - - 50 MPa -BT cao : 60 - 100 MPa - 32 - - 5 ... - tailieumienphi.vn