Đáp án:
Chương VII:
\(\begin{array}{l}
1.C\\
2.C\\
3.B\\
4.B.\Delta l = {l_0}\alpha \Delta t\\
5.C\\
6.C\\
7.D\\
8.D.0,22mm\\
\Delta l = {l_0}\alpha \Delta t = {1.11.10^{ - 6}}(40 - 20) = 2,{2.10^{ - 4}}m = 0,22mm\\
9.C.3,6mm\\
\Delta l = {l_0}\alpha \Delta t = {10.12.10^{ - 6}}(40 - 10) = 3,{6.10^{ - 3}}m = 3,6mm\\
10.B.F = 0,002N\\
F = \sigma l = 0,04.0,05 = 0,002N\\
11.A.0,02N\\
{F_c} = F - P = 0,06 - 0,04 = 0,02N
\end{array}\)
Chương VI:
\(\begin{array}{l}
1.B\\
2A.Q = mc\Delta t\\
3.B\\
4.D\\
5.A\\
\Delta U = A + Q \Rightarrow Q = \Delta U - A = 20 - 100 = - 80J\\
6.C\\
7.B\\
Q = 100J\\
A = - 70J\\
\Delta U = A + Q = - 70 + 100 = 30J\\
8.C.20\% \\
H = \dfrac{{{Q_1} - {Q_2}}}{{{Q_1}}} = \dfrac{{400 - 320}}{{400}} = 20\% \\
9.D\\
10.C.33,{44.10^4}J\\
Q = mc\Delta t = 1.4180.(100 - 20) = 33,{44.10^4}J\\
11.C.7,{66.10^3}kg/{m^3}\\
\beta = 3\alpha = {3.12.10^{ - 6}} = {36.10^{ - 6}}{K^{ - 1}}\\
V = {V_0}(1 + \beta \Delta t) \Rightarrow \frac{m}{D} = \dfrac{m}{{{D_0}}}(1 + \beta \Delta t)\\
\Rightarrow D = \dfrac{{{D_0}}}{{1 + \beta \Delta t}} = \dfrac{{7,{{8.10}^3}}}{{1 + {{36.10}^{ - 6}}.500}} = 7,{66.10^3}kg/{m^3}
\end{array}\)
