Đáp án:

\(\begin{array}{l}
1.D.1,{137.10^5}Pa\\
{T_1} = 20 + 273 = 293^\circ K\\
{T_2} = 60 + 273 = 333^\circ K\\
\dfrac{{{p_1}}}{{{T_1}}} = \dfrac{{{p_2}}}{{{T_2}}} \Rightarrow \dfrac{{{{10}^5}}}{{293}} = \dfrac{{{p_2}}}{{333}}\\
 \Rightarrow {p_2} = 1,{137.10^5}Pa\\
2.B\\
3.D.\dfrac{v}{3}\\
p = p'\\
mv = (m + 2m)v' \Rightarrow v' = \dfrac{{mv}}{{3m}} = \dfrac{v}{3}\\
4.A.20s\\
A = Ph = mgh = 10.1000.30 = 300000J\\
P = \dfrac{A}{t} \Rightarrow t = \dfrac{A}{P} = \dfrac{{300000}}{{15000}} = 20s\\
5.B.13,5kg.m/s\\
{p_1} = {m_1}{v_1} = 1.6 = 6kg.m/s\\
{p_2} = {m_2}{v_2} = 2.4 = 8kg.m/s\\
p = \sqrt {p_1^2 + p_2^2 + 2{p_1}{p_2}\cos 30}  = \sqrt {{6^2} + {8^2} + 2.6.8.\cos 30}  = 13,5kg.m/s\\
6.A.3J\\
W = {W_{d\max }} = \dfrac{1}{2}mv_{\max }^2 = \dfrac{1}{2}0,{1.10^2} = 5J\\
W = {W_{t\max }} = mg{h_{\max }} \Rightarrow 5 = 0,1.10.{h_{\max }}\\
 \Rightarrow {h_{\max }} = 5m\\
 \Rightarrow h = {h_{\max }} - (s - {h_{\max }}) = 5 - (8 - 5) = 2m\\
{W_t} = mgh = 0,1.10.2 = 2J\\
{W_d} = W - {W_t} = 5 - 2 = 3J\\
7.A.1kgm/s\\
p = |{p_1} - {p_2}| = |{m_1}{v_1} + {m_2}{v_2}| = |1,5.2 - 2.2| = 1kgm/s\\
8.C.A = 1320J,P = 330W\\
s = \dfrac{1}{2}a{t^2} \Rightarrow a = \dfrac{{2s}}{{{a^2}}} = \dfrac{{2.8}}{{{4^2}}} = 1m/{s^2}\\
F - P = ma \Rightarrow F = ma + mg = 15.1 + 15.10 = 165N\\
A = Fs = 165.8 = 1320J\\
P = \dfrac{A}{t} = \dfrac{{1320}}{4} = 330W\\
9.B.5kg\\
p = p'\\
{m_1}{v_1} = {m_2}{v_2} - {m_1}{v_1}\\
 \Rightarrow 2.4 = 2{m_2} - 2.1\\
 \Rightarrow {m_2} = 5kg\\
10.C\\
11.C{.8.10^3}N\\
a = \dfrac{{v_2^2 - v_1^2}}{{2s}} = \dfrac{{{{100}^2} - {{300}^2}}}{{2.0,05}} =  - {800.10^3}\\
{F_c} =  - ma =  - 0,01.( - {800.10^3}) = {8.10^3}N\\
12.B\\
13.A. - 6m/s\\
p = p' \Rightarrow 0 = {m_1}{v_1} + {m_2}{v_2}\\
 \Rightarrow 10.600 + 1000{v_2} = 0\\
 \Rightarrow {v_2} =  - 6m/s\\
14.C.6,{68.10^{ - 27}}kg\\
p = p'\\
{m_p}{v_p} = {m_a}{v_a}' - {m_p}{v_p}'\\
 \Rightarrow 1,{67.10^{ - 27}}{.10^7} = {m_a}{.4.10^6} - 1,{67.10^{ - 27}}{.6.10^6}\\
 \Rightarrow {m_a} = 6,{68.10^{ - 27}}kg\\
15.A\\
16.B.10l\\
{T_1} = 27 + 273 = 300^\circ K\\
{T_2} = 227 + 273 = 500^\circ K\\
\dfrac{{{V_1}}}{{{T_1}}} = \dfrac{{{V_2}}}{{{T_2}}} \Rightarrow \dfrac{6}{{300}} = \dfrac{{{V_2}}}{{500}}\\
 \Rightarrow {V_2} = 10l\\
17.C{.2.10^5}J\\
{W_d} = \dfrac{1}{2}m{v^2} = \dfrac{1}{2}{.1000.20^2} = {2.10^5}J\\
18.D{.24.10^4}J\\
s = vt = 10.20.60 = 12000m\\
A = Fs\cos 60 = 40.12000.\cos 60 = {24.10^4}J\\
19.D\\
20.A.40kgm/s\\
{W_t} = {W_d} \Rightarrow mgh = \dfrac{1}{2}m{v^2}\\
 \Rightarrow 10.20 = \dfrac{1}{2}{v^2} \Rightarrow v = 20m/s\\
\Delta p = m(v - {v_0}) = 2.(20 - 0) = 40kgm/s\\
21.D\\
22.B.15l\\
{T_1} = 273 + 273 = 546^\circ K\\
{T_2} = 546 + 273 = 819^\circ K\\
\dfrac{{{V_1}}}{{{T_1}}} = \dfrac{{{V_2}}}{{{T_2}}} \Rightarrow \dfrac{{10}}{{546}} = \dfrac{{{V_2}}}{{819}}\\
 \Rightarrow {V_2} = 15l\\
23.C.5,{2.10^5}Pa\\
{T_1} = 30 + 273 = 303^\circ K\\
{T_2} = 40 + 273 = 313^\circ K\\
\dfrac{{{p_1}{V_1}}}{{{T_1}}} = \dfrac{{{p_2}{V_2}}}{{{T_2}}} \Rightarrow \dfrac{{{{10}^5}.150}}{{303}} = \dfrac{{{p_2}30}}{{313}}\\
 \Rightarrow {p_2} = 5,{2.10^5}Pa\\
24.D.1,{944.10^5}J\\
s = \dfrac{1}{2}a{t^2} \Rightarrow a = \dfrac{{2s}}{{{t^2}}} = \dfrac{{2.10}}{{{5^2}}} = 0,8m/{s^2}\\
{s_4} = \dfrac{1}{2}a{t_4}^2 = \dfrac{1}{2}.0,{8.4^2} = 6,4m\\
s' = s - {s_4} = 10 - 6,4 = 3,6m\\
F - P = ma \Rightarrow F = ma + mg = 5000.0,8 + 5000.10 = 54000N\\
A = Fs = 54000.3,6 = 1,{944.10^5}J\\
25.D\\
26.C.2,4m\\
W = {W_{t\max }} \Rightarrow mgh + \dfrac{1}{2}m{v^2} = mg{h_{\max }}\\
 \Rightarrow 10.1,6 + \dfrac{1}{2}{.4^2} = 10{h_{\max }}\\
 \Rightarrow {h_{\max }} = 2,4m\\
27.B\\
28.A.0,9m\\
{W_{d\max }} = {W_{t\max }} \Rightarrow \dfrac{1}{2}mv_{\max }^2 = mg{h_{\max }}\\
 \Rightarrow \dfrac{1}{2}{.6^2} = 10{h_{\max }} \Rightarrow {h_{\max }} = 1,8m\\
{W_t} = {W_d} = \dfrac{W}{2} = \dfrac{{{W_{d\max }}}}{2} \Rightarrow mgh = \dfrac{{mg{h_{\max }}}}{2}\\
 \Rightarrow h = \dfrac{{{h_{\max }}}}{2} = \dfrac{{1,8}}{2} = 0,9m\\
29.B.207^\circ C\\
{T_1} = 47 + 273 = 320^\circ K\\
\dfrac{{{p_1}{V_1}}}{{{T_1}}} = \dfrac{{{p_2}{V_2}}}{{{T_2}}} \Rightarrow \dfrac{{1.2}}{{320}} = \dfrac{{15.0,2}}{{{T_2}}}\\
 \Rightarrow {T_2} = 480^\circ K\\
{t_2} = {T_2} - 273 = 480 - 273 = 207^\circ C\\
30.\\
{W_t} = \dfrac{1}{2}k\Delta {l^2}
\end{array}\)