Các đáp án câu khoanh được sắp xếp từ câu 41->49 

$C$

$D$

$C$

$A$

$B$

$A$

$C$

$B$

$A$

$NOCOPY$

Đáp án:

\(\begin{array}{l}
41.C\\
{T_1} = 27 + 273 = 300^\circ K\\
{T_2} = 17 + 273 = 290^\circ K\\
\frac{{{p_1}{V_1}}}{{{T_1}}} = \dfrac{{{p_2}{V_2}}}{{{T_2}}} \Rightarrow {V_2} = \dfrac{{{p_1}{V_1}{T_2}}}{{{T_1}{p_2}}} = \dfrac{{750.40.290}}{{300.720}} = 40,3c{m^3}\\
42.D\\
{T_1} = 20 + 273 = 293^\circ K\\
{T_2} = 40 + 273 = 313^\circ K\\
\dfrac{{{p_1}}}{{{T_1}}} = \dfrac{{{p_2}}}{{{T_2}}} \Rightarrow {p_2} = \dfrac{{{p_1}{T_2}}}{{{T_1}}} = \dfrac{{{{10}^5}.313}}{{293}} = 1,{07.10^5}Pa\\
43.C\\
{p_1}{V_1} = {p_2}{V_2} \Rightarrow {V_2} = \dfrac{{{p_1}{V_1}}}{{{p_2}}} = \dfrac{{1.2}}{{3,5}} = 0,57{m^3}\\
44.A\\
45.B\\
p = mv \Rightarrow v = \dfrac{p}{m} = \dfrac{8}{{1,6}} = 5m/s\\
46.A\\
{{\rm{W}}_t} = mgh \Rightarrow h = \dfrac{{{{\rm{W}}_t}}}{{mg}} = \dfrac{{300}}{{1,5.10}} = 20m\\
47.C\\
48.B\\
{{\rm{W}}_d} = \dfrac{1}{2}m{v^2} \Rightarrow v = \sqrt {\dfrac{{2{W_d}}}{m}}  = \sqrt {\frac{{2.10}}{{0,2}}}  = 10m/s\\
49.A\\
{p_1}{V_1} = {p_2}{V_2} \Rightarrow {p_2} = \dfrac{{{p_1}{V_1}}}{{{V_2}}} = \dfrac{{{p_1}.10}}{5} = 2{p_2}
\end{array}\)