1.
![^{251}_{98}Cf = ^{211}_{85}At + k*^{4}_{2}He + m*^{0}_{-1}e \\ \\ \left \{ {{251=211+4k} \atop {98=85+2k-m}} \right. \\ \left \{ {{k=10} \atop {m=7}} \right \\ \\ ^{251}_{98}Cf = ^{211}_{85}At + 10^{4}_{2}He + 7^{0}_{-1}e ](https://tex.z-dn.net/?f=%5E%7B251%7D_%7B98%7DCf+%3D+%5E%7B211%7D_%7B85%7DAt+%2B+k%2A%5E%7B4%7D_%7B2%7DHe+%2B+m%2A%5E%7B0%7D_%7B-1%7De+%5C%5C+%5C%5C++%5Cleft+%5C%7B+%7B%7B251%3D211%2B4k%7D+%5Catop+%7B98%3D85%2B2k-m%7D%7D+%5Cright.++%5C%5C++%5Cleft+%5C%7B+%7B%7Bk%3D10%7D+%5Catop+%7Bm%3D7%7D%7D+%5Cright+%5C%5C+%5C%5C++%5E%7B251%7D_%7B98%7DCf+%3D+%5E%7B211%7D_%7B85%7DAt+%2B+10%5E%7B4%7D_%7B2%7DHe+%2B+7%5E%7B0%7D_%7B-1%7De%0A)
Т. е. происходит 10 α-распадов и 7 β-распадов.
![^{254}_{99}Es = ^{222}_{87}Fr + k*^{4}_{2}He + m*^{0}_{-1}e \\ \\ \left \{ {{254=222+4k} \atop {99=87+2k-m}} \right. \\ \left \{ {{k=8} \atop {m=4}} \right \\ \\^{254}_{99}Es = ^{222}_{87}Fr + 8^{4}_{2}He + 4^{0}_{-1}e](https://tex.z-dn.net/?f=%5E%7B254%7D_%7B99%7DEs+%3D+%5E%7B222%7D_%7B87%7DFr+%2B+k%2A%5E%7B4%7D_%7B2%7DHe+%2B+m%2A%5E%7B0%7D_%7B-1%7De+%5C%5C+%5C%5C++%5Cleft+%5C%7B+%7B%7B254%3D222%2B4k%7D+%5Catop+%7B99%3D87%2B2k-m%7D%7D+%5Cright.++%5C%5C++%5Cleft+%5C%7B+%7B%7Bk%3D8%7D+%5Catop+%7Bm%3D4%7D%7D+%5Cright+%5C%5C+%5C%5C%5E%7B254%7D_%7B99%7DEs+%3D+%5E%7B222%7D_%7B87%7DFr+%2B+8%5E%7B4%7D_%7B2%7DHe+%2B+4%5E%7B0%7D_%7B-1%7De)
Т. е. происходит 8 α-распадов и 4 β-распада.
2.
![^{16}_{8}O + ^{1}_{0}n = ^{14}_{7}N + ^{k}_{m}E \\ \\ \left \{ {{16+1=14+k} \atop {8+0=7+m}} \right. \\ \left \{ {{k=3} \atop {m=1}} \right \\ \\^{3}_{1}E = ^{3}_{1}H \\ \\ ^{16}_{8}O + ^{1}_{0}n = ^{14}_{7}N +^{3}_{1}H](https://tex.z-dn.net/?f=%5E%7B16%7D_%7B8%7DO+%2B+%5E%7B1%7D_%7B0%7Dn+%3D+%5E%7B14%7D_%7B7%7DN+%2B+%5E%7Bk%7D_%7Bm%7DE+%5C%5C+%5C%5C+%5Cleft+%5C%7B+%7B%7B16%2B1%3D14%2Bk%7D+%5Catop+%7B8%2B0%3D7%2Bm%7D%7D+%5Cright.+%5C%5C+%5Cleft+%5C%7B+%7B%7Bk%3D3%7D+%5Catop+%7Bm%3D1%7D%7D+%5Cright+%5C%5C+%5C%5C%5E%7B3%7D_%7B1%7DE+%3D+%5E%7B3%7D_%7B1%7DH+%5C%5C+%5C%5C++%5E%7B16%7D_%7B8%7DO+%2B+%5E%7B1%7D_%7B0%7Dn+%3D+%5E%7B14%7D_%7B7%7DN+%2B%5E%7B3%7D_%7B1%7DH)
![^{15}_{7}N + ^{4}_{2} \alpha = ^{12}_{6}C + ^{k}_{m}E \\ \\ \left \{ {{15+4=12+k} \atop {7+2=6+m}} \right. \\ \left \{ {{k=7} \atop {m=3}} \right \\ \\^{7}_{3}E = ^{7}_{3}Li \\ \\ ^{15}_{7}N + ^{4}_{2} \alpha = ^{12}_{6}C + ^{7}_{3}Li](https://tex.z-dn.net/?f=%5E%7B15%7D_%7B7%7DN+%2B+%5E%7B4%7D_%7B2%7D+%5Calpha+%3D+%5E%7B12%7D_%7B6%7DC+%2B+%5E%7Bk%7D_%7Bm%7DE+%5C%5C+%5C%5C+%5Cleft+%5C%7B+%7B%7B15%2B4%3D12%2Bk%7D+%5Catop+%7B7%2B2%3D6%2Bm%7D%7D+%5Cright.+%5C%5C+%5Cleft+%5C%7B+%7B%7Bk%3D7%7D+%5Catop+%7Bm%3D3%7D%7D+%5Cright+%5C%5C+%5C%5C%5E%7B7%7D_%7B3%7DE+%3D+%5E%7B7%7D_%7B3%7DLi+%5C%5C+%5C%5C++%5E%7B15%7D_%7B7%7DN+%2B+%5E%7B4%7D_%7B2%7D+%5Calpha+%3D+%5E%7B12%7D_%7B6%7DC+%2B+%5E%7B7%7D_%7B3%7DLi)
3.
![^{2}_{1}H + ^{3}_{1}H= ^{4}_{2}He + ^{k}_{m}E \\ \\ \left \{ {{2+3=4+k} \atop {1+1=2+m}} \right. \\ \left \{ {{k=1} \atop {m=0}} \right \\ \\^{1}_{0}E = ^{1}_{0}n \\ \\^{2}_{1}H + ^{3}_{1}H= ^{4}_{2}He + ^{1}_{0}n \\ \\ \\^{6}_{3}Li + ^{1}_{1}H= ^{4}_{2}He + ^{k}_{m}E \\ \\ \left \{ {{6+1=4+k} \atop {3+1=2+m}} \right. \\ \left \{ {{k=3} \atop {m=2}} \right \\ \\^{3}_{2}E = ^{3}_{2}He \\ \\^{6}_{3}Li + ^{1}_{1}H= ^{4}_{2}He +^{3}_{2}He](https://tex.z-dn.net/?f=%5E%7B2%7D_%7B1%7DH+%2B+%5E%7B3%7D_%7B1%7DH%3D+%5E%7B4%7D_%7B2%7DHe+%2B+%5E%7Bk%7D_%7Bm%7DE+%5C%5C+%5C%5C+%5Cleft+%5C%7B+%7B%7B2%2B3%3D4%2Bk%7D+%5Catop+%7B1%2B1%3D2%2Bm%7D%7D+%5Cright.+%5C%5C+%5Cleft+%5C%7B+%7B%7Bk%3D1%7D+%5Catop+%7Bm%3D0%7D%7D+%5Cright+%5C%5C+%5C%5C%5E%7B1%7D_%7B0%7DE+%3D+%5E%7B1%7D_%7B0%7Dn+%5C%5C+%5C%5C%5E%7B2%7D_%7B1%7DH+%2B+%5E%7B3%7D_%7B1%7DH%3D+%5E%7B4%7D_%7B2%7DHe+%2B+%5E%7B1%7D_%7B0%7Dn+%5C%5C++%5C%5C++%5C%5C%5E%7B6%7D_%7B3%7DLi+%2B+%5E%7B1%7D_%7B1%7DH%3D+%5E%7B4%7D_%7B2%7DHe+%2B+%5E%7Bk%7D_%7Bm%7DE+%5C%5C+%5C%5C+%5Cleft+%5C%7B+%7B%7B6%2B1%3D4%2Bk%7D+%5Catop+%7B3%2B1%3D2%2Bm%7D%7D+%5Cright.+%5C%5C+%5Cleft+%5C%7B+%7B%7Bk%3D3%7D+%5Catop+%7Bm%3D2%7D%7D+%5Cright+%5C%5C+%5C%5C%5E%7B3%7D_%7B2%7DE+%3D+%5E%7B3%7D_%7B2%7DHe+%5C%5C+%5C%5C%5E%7B6%7D_%7B3%7DLi+%2B+%5E%7B1%7D_%7B1%7DH%3D+%5E%7B4%7D_%7B2%7DHe+%2B%5E%7B3%7D_%7B2%7DHe+)
Находить по формуле: U=I*R
U=0.5A * 20Ом=10В
Ответ: 10В
Дано:
а = 1 см = 0,01 м
b = 2,5 см = 0,025 м.
с = 0,8 см = 0,008 м
m = 3,2 г = 0,0032 кг
ρ - ?
Решение:
ρ = m/V
V = а *b *с
V = 0,01*0,025*0,008 = 0,000002 м^3
ρ = 0,0032 / 0,000002 = 1600 кг/м^3.
V=s1/t1=20/4=5 м/с
Где s-Путь. t1 время в 4 сек.
а=v/t= 5/10=0.5 м/c2
S2=V^2/(2a)=5^2/2*0.5= 25/1= 25 м