Đề thi thử tốt nghiệp THPT môn Vật lí Sở Hậu Giang có đáp án

${Q_{dd}} = qm = 2{,}7 \cdot {10^7}.200 = 5,{4.10^9}\;{\rm{J}}$

${Q_t} = H{Q_{dd}} = 0{,}6.5,{4.10^9} = 3,{24.10^9}\;{\rm{J}} \Rightarrow$ a) Đúng; b) Đúng

${Q_t} = {m_t}c\Delta t + {m_t}\lambda \Rightarrow 3{,}24 \cdot {10^9} = {m_t} \cdot 460 \cdot (1400 - 20) + {m_t} \cdot 83{,}7 \cdot {10^3} \Rightarrow {m_t} \approx 4509{,}4\;{\rm{kg}}$

$\Rightarrow$ c) Sai; d) Sai

$e = \left| {\frac{{\Delta \phi }}{{\Delta t}}} \right| = \left| {\frac{{N \cdot \Delta B \cdot S}}{{\Delta t}}} \right| = 12 \cdot 5 \cdot 40 \cdot {10^{ - 4}} = 0{,}24V$

$\Rightarrow$ a) Đúng; b) Đúng; c) Đúng; d) Sai

a) Đúng

${N_B} = {A_B} - {Z_B} = 204 - 80 = 124 > {N_A} = 122 \Rightarrow$b) Sai

$\left\{ {\begin{array}{llllllllllllllllllll}{\frac{{\Delta {m_A}}}{{{A_A}}} = \frac{{1{,}71228}}{{202}} \approx 8{,}4766 \cdot {{10}^{ - 3}}}\\{\frac{{\Delta {m_B}}}{{{A_B}}} = \frac{{1{,}72675}}{{204}} \approx 8,{{464.10}^{ - 3}}}\end{array} \Rightarrow \frac{{\Delta {m_A}}}{{{A_A}}} > \frac{{\Delta {m_B}}}{{{A_B}}} \Rightarrow {W_{lkA}} > {W_{lkrB}} \Rightarrow } \right.$ A bên hơn B $\Rightarrow$ c) Sai

${Z_A} = {A_A} - {N_A} = 202 - 122 = 80 = {Z_B} \Rightarrow$d) Đúng

Đẳng áp $\frac{V}{T} =$ const $\Rightarrow$ a) Đúng

$\frac{{{V_0}}}{{{T_0}}} = \frac{{{V_1}}}{{{T_1}}} \Rightarrow \frac{{{V_0}}}{{20 + 273}} = \frac{{{V_1}}}{{50 + 273}} \Rightarrow \frac{{{V_1}}}{{{V_0}}} \approx 1{,}102 > 1{,}101 \Rightarrow$b) Đúng

Nếu T giảm thì V giảm $\Rightarrow$ c) Đúng

$\frac{{{V_0}}}{{{T_0}}} = \frac{{{V_2}}}{{{T_2}}} \Rightarrow \frac{{{V_0}}}{{20 + 273}} = \frac{{{V_2}}}{{40 + 273}} \Rightarrow \frac{{{V_2}}}{{{V_0}}} \approx 1{,}068 \Rightarrow$d) Sai

$\left| {\frac{{\Delta \phi }}{{\Delta t}}} \right| = \left| {\frac{{\Delta B \cdot S}}{{\Delta t}}} \right| = \frac{{(9 - 3) \cdot {{10}^{ - 3}} \cdot 20 \cdot {{10}^{ - 4}}}}{{0{,}2}} = 60 \cdot {10^{ - 6}}\;{\rm{Wb}}/{\rm{s}} = 60\mu \;{\rm{Wb}}/{\rm{s}}$

Trả lời ngắn: 60

$e = - \frac{{\Delta \phi }}{{\Delta t}} = - \frac{{\Delta B \cdot S}}{{\Delta t}} = \frac{{9 \cdot {{10}^{ - 3}} \cdot 20 \cdot {{10}^{ - 4}}}}{{1 - 0{,}4}} = 3 \cdot {10^{ - 5}}\;{\rm{V}}$

$R = \frac{e}{i} = \frac{{3 \cdot {{10}^{ - 5}}}}{{0{,}2 \cdot {{10}^{ - 3}}}} = 0{,}15\Omega$

Trả lời ngắn: 0,15

$Q = \left( {{m_{nh}}{c_{nh}} + {m_n}{c_n}} \right)\Delta t = (0{,}5 \cdot 880 + 2 \cdot 4200) \cdot (100 - 25) = 663000\;{\rm{J}} = 663\;{\rm{kJ}}$

Trả lời ngắn: 663

$A = \frac{Q}{H} = \frac{{663000}}{{0{,}75}} = 884000\;{\rm{J}}$

$P = \frac{A}{t} = \frac{{884000}}{{400}} = 2210\;{\rm{W}}$

Trả lời ngắn: 2210

$_{92}^{235}U + _0^1n \to _{54}^{140}X + _{38}^{94}Sr + 2_0^1n$

Hạt nhân $X$ có $N = A - Z = 140 - 54 = 86$ neutron

Trả lời ngắn: 86

$n = \frac{m}{M} = \frac{{1000}}{{235}} = \frac{{200}}{{47}}\;{\rm{mol}}$

$N = n{N_A} = \frac{{200}}{{47}} \cdot 6{,}02 \cdot {10^{23}} \approx 2{,}562 \cdot {10^{24}}$

$Q = N\Delta E = 2{,}562 \cdot {10^{24}} \cdot 203 \approx 5{,}2 \cdot {10^{26}}{\rm{MeV}}$

Trả lời ngắn: 5,2