Đề tham khảo Tốt nghiệp THPT Vật Lí 2026-2027 (Đề 29)

$\omega =2\pi f=2\pi .350=700\pi (\text{r}\text{a}\text{d}/\text{s})$

$E{}_0=NBS\omega =500.0,05.0,008.700\pi \approx 440 \text{V}.$ Đáp án đúng D

$W{}_{lkr}=\frac{W{}_{lk}}{A}=\frac{2,56}{2}=1,28\text{M}\text{e}\text{V}/\text{n}\text{u}\text{c}\text{l}\text{e}\text{o}\text{n}\text{.}$Đáp án đúng D

Theo quy tắc ra bắc vào nam (ra N vào S) thì đường sức từ hướng sang phải. Áp dụng quy tắc bàn tay trái được lực từ theo hướng A. Đáp án đúng A

Đáp án đúng A

Máy biến áp không làm thay đổi tần số. Đáp án đúng C

$F=BI\mathcal{l}sin\alpha$. Đáp án đúng D

Đáp án đúng A

${}_0^{1}n+{}_{92}^{235}U\to {}_{39}^{95}Y+{}_{53}^{138}I+3{}_0^{1}n$. Đáp án đúng B

$-8<-2<60$ là chất 3 . Đáp án đúng A

a) Đúng. Lực điện trường làm tăng tốc hạt.

b) Đúng.

$F=ma{}_{ht}\Rightarrow qvB=m\cdot \frac{v{}^2}{R}\Rightarrow R=\frac{mv}{qB}\Rightarrow R{}_1=\frac{3,31\cdot 10{}^{-27}\cdot 3,2\cdot 10{}^6}{1,6\cdot 10{}^{-19}\cdot 1,6}=0,041375 \text{m}=4,1375 \text{c}\text{m}$

c) Đúng. Định lý động năng $\Delta W{}_d=A\Rightarrow \frac{1}{2}mv{}_2^2-\frac{1}{2}mv{}_1^2=qU$

$\Rightarrow \frac{1}{2}\cdot 3,31\cdot 10{}^{-27}\cdot v{}_2^2-\frac{1}{2}\cdot 3,31\cdot 10{}^{-27}\cdot \left(3,2\cdot 10{}^6\right){}^2=1,6\cdot 10{}^{-19}\cdot 100\cdot 10{}^3\Rightarrow v{}_2\approx 4,4618\cdot 10{}^6 \text{m}/\text{s}$

$R{}_2=\frac{mv{}_2}{qB}=\frac{3,31\cdot 10{}^{-27}\cdot 4,4618\cdot 10{}^6}{1,6\cdot 10{}^{-19}\cdot 1,6}\approx 0,058 \text{m}=5,8 \text{c}\text{m}$

d) Sai. $R{}_{max}=\frac{mv{}_{max}}{qB}\Rightarrow 0,5=\frac{3,31\cdot 10{}^{-27}\cdot v{}_{max}}{1,6\cdot 10{}^{-19}\cdot 1,6}\Rightarrow v{}_{max}\approx 38,67\cdot 10{}^6 \text{m}/\text{s}$

$\frac{1}{2}mv{}_{max{}^2}-\frac{1}{2}mv{}_1^2=NqU$

$\Rightarrow \frac{1}{2}\cdot 3,31\cdot 10{}^{-27}\cdot \left(38,67\cdot 10{}^6\right){}^2-\frac{1}{2}\cdot 3,31\cdot 10{}^{-27}\cdot \left(3,2\cdot 10{}^6\right){}^2=N\cdot 1,6\cdot 10{}^{-19}\cdot 100\cdot 10{}^3$$\Rightarrow N\approx 153,6$

a) Đúng. $Q=Pt=800\cdot (2+5+5)\cdot 60=576000 \text{J}=576 \text{k}\text{J}$

$\{\begin{array}{c}P\Delta t{}_1=m{}_{0}c\left(t{}_1-t{}_0\right)\\ P\Delta t{}_2=m{}_{0}c\left(t{}_2-t{}_1\right)+m{}_{t}c\left(t{}_2-t{}_x\right)\\ P\Delta t{}_3=\left(m{}_0+m{}_t\right)c\left(t{}_s-t{}_2\right)\end{array}$

\Rightarrow \frac{P}{c}=\frac{m{}_0}{4}=\frac{m{}_0}{15}+\frac{m{}_t\left(70-t{}_x\right)}{300}=\frac{m{}_0+m{}_t}{10}\Rightarrow m{}_t=1,5m{}_0\to t{}_x\approx 33,3{}^{°}C\Rightarrowb) Sai; c) Sai

d) Sai. $P=\frac{m{}_{0}c}{4}=\frac{1.4200}{4}=1050 \text{W}$

a) Đúng. ${}_{53}^{131}\text{I}\to {}_{54}^{131}\text{X}\text{e}+{}_{-1}^{0}e$

b) Sai. $\lambda =\frac{ln2}{T}=\frac{ln2}{8,04\cdot 24\cdot 60\cdot 60}\approx 9,98\cdot 10{}^{-7} \text{s}{}^{-1}$

c) Sai. $H{}_0=\lambda N{}_0\Rightarrow 5,2\cdot 10{}^8=9,98\cdot 10{}^{-7}\cdot N{}_0\Rightarrow N{}_0=5,21\cdot 10{}^{14}$

$n{}_0=\frac{N{}_0}{N{}_A}=\frac{5,21\cdot 10{}^{14}}{6,02\cdot 10{}^{23}}\approx 8,65\cdot 10{}^{-10} \text{m}\text{o}\text{l}$

$m{}_0=n{}_{0}M=8,65\cdot 10{}^{-10}\cdot 131\approx 0,113\cdot 10{}^{-6}g=0,113\mu g$

d) Sai. $H=0,85H{}_0\cdot 2{}^{\frac{-t}{T}}=0,85\cdot 5,2\cdot 10{}^8\cdot 2{}^{\frac{-48}{8,04\cdot 24}}\approx 3,72\cdot 10{}^{8}Bq$

a) Đúng

b) Đúng. $p{}_1=p{}_0+\frac{m{}_{gh}g}{S}=10{}^5+\frac{5.10}{20.10{}^{-4}}=1,25.10{}^5 \text{P}\text{a}$

c) Sai. $p{}_2=p{}_0+\frac{\left(m{}_{gh}+m{}_{hs}\right)g}{S}=10{}^5+\frac{(5+50)\cdot 10}{20\cdot 10{}^{-4}}=3,75\cdot 10{}^5 \text{P}\text{a}$

d) Sai. $pV=$ const $\Rightarrow p{}_{1}S\mathcal{l}{}_1=p{}_{2}S\mathcal{l}{}_2\Rightarrow 1,25\cdot 10{}^5\cdot \mathcal{l}{}_1=3,75\cdot 10{}^5\cdot \left({\mathcal{l}}_1-15\right)\Rightarrow \mathcal{l}{}_1=22,5 \text{c}\text{m}$

$\frac{pV}{T}=\text{c}\text{o}\text{n}\text{s}\text{t}\Rightarrow \frac{1\cdot 2,5}{27+273}=\frac{(1+17)\cdot 0,25}{T}\Rightarrow T=540 \text{K}=267{}^{°}\text{C}$

Đáp án: 267

$p{}_{1}V{}_1=p{}_{2}V{}_2\Rightarrow 3.0,5=p{}_2.0,2\Rightarrow p{}_2=7,5 \text{a}\text{t}\text{m}$

Đáp án: 7,5

${\Phi }_0=NBS=500.5\cdot 10{}^{-5}\cdot 20.10{}^{-4}=5.10{}^{-5} \text{W}\text{b}=50\mu \text{W}\text{b}$

Đáp án: 50

$e=\left|\frac{\Delta \Phi }{\Delta t}\right|=\frac{5.10{}^{-5}}{0,01}=5.10{}^{-3} \text{V}=5\text{m}\text{V}$

Đáp án: 5

$N{}_0=nN{}_A=0,5\cdot 6,02\cdot 10{}^{23}=3,01\cdot 10{}^{23}$

$H{}_0=\lambda N{}_0=\frac{ln2}{T}\cdot N{}_0=\frac{ln2}{1,251\cdot 10{}^9\cdot 365\cdot 24\cdot 60\cdot 60}\cdot 3,01\cdot 10{}^{23}\approx 5,29\cdot 10{}^{6}Bq$

$P=H{}_{0}E{}_1=5,29\cdot 10{}^6\cdot 4,6\cdot 1,6\cdot 10{}^{-13}\approx 3,89\cdot 10{}^{-6} \text{J}/\text{s}=3,89\mu \text{J}/\text{s}$

Đáp án: 3,89

$\frac{\Delta N}{N}=\frac{N{}_0\left(1-2{}^{\frac{-t}{T}}\right)}{N{}_0\cdot 2{}^{\frac{-t}{T}}}=2{}^{\frac{t}{T}}-1\Rightarrow 10,3=2{}^{\frac{t}{1,251}}-1\Rightarrow t\approx 4,4$tỉ năm

Đáp án: 4,4