E=E1-E2=kq1/(r1)^2 - kq2/(r2)^2=
=9*10^9*6.7*10^-9/(3*10^-2)- 9*10^9*13.3&10^-9/(4*10^-2)= ....
ну сам решай с помошю калькулятора
Дано:
t
![_{l}](https://tex.z-dn.net/?f=+_%7Bl%7D+)
и t
![_{al}](https://tex.z-dn.net/?f=+_%7Bal%7D+)
=-200C
m
![_{l}](https://tex.z-dn.net/?f=+_%7Bl%7D+)
=2 кг
m
![_{al}](https://tex.z-dn.net/?f=+_%7Bal%7D+)
=0.5 кг
с
![_{al}](https://tex.z-dn.net/?f=+_%7Bal%7D+)
=930 <span>Дж/(кг·С)
с</span>
![_{l}](https://tex.z-dn.net/?f=+_%7Bl%7D+)
=2100 Дж/(кг·С)
c
![_{B}](https://tex.z-dn.net/?f=+_%7BB%7D+)
=4200 Дж/(кг·С)
λ
![_{l}](https://tex.z-dn.net/?f=+_%7Bl%7D+)
=3.3×10⁵ <span>Дж/кг
</span>L
![_{B}](https://tex.z-dn.net/?f=+_%7BB%7D+)
=2.26×10⁶ Дж/кг
q=1.3×10⁷ <span>Дж/кг
</span>η=15%<span>
Найти:m=?
Решение:
Q</span>₁+Q₂+Q₃+Q₄+Q₅=Q₆
Q₁=с
![_{al}](https://tex.z-dn.net/?f=+_%7Bal%7D+)
m
![_{al}](https://tex.z-dn.net/?f=+_%7Bal%7D+)
Δtt
![_{al}](https://tex.z-dn.net/?f=+_%7Bal%7D+)
Q₂=с
![_{l}](https://tex.z-dn.net/?f=+_%7Bl%7D+)
m
![_{l}](https://tex.z-dn.net/?f=+_%7Bl%7D+)
Δt
![_{l}](https://tex.z-dn.net/?f=+_%7Bl%7D+)
Q₃=λ
![_{l}](https://tex.z-dn.net/?f=+_%7Bl%7D+)
m
![_{l}](https://tex.z-dn.net/?f=+_%7Bl%7D+)
Q₄=m
![_{B}](https://tex.z-dn.net/?f=+_%7BB%7D+)
c
![_{B}](https://tex.z-dn.net/?f=+_%7BB%7D+)
Δt
![_{B}](https://tex.z-dn.net/?f=+_%7BB%7D+)
Q₅=L
![_{B}](https://tex.z-dn.net/?f=+_%7BB%7D+)
m
![_{B}](https://tex.z-dn.net/?f=+_%7BB%7D+)
Q₆=qm
m=
![\frac{Q_{1} +Q_{2}+Q_{3}+Q_{4}+Q_{5}}{q}](https://tex.z-dn.net/?f=+%5Cfrac%7BQ_%7B1%7D+%2BQ_%7B2%7D%2BQ_%7B3%7D%2BQ_%7B4%7D%2BQ_%7B5%7D%7D%7Bq%7D+)
m=
![\frac{930*0.5*t+2100*2*200+3.3*10^{5}+4200*2*100+2.26*10^{6}*2 }{1.3*10^{7} }](https://tex.z-dn.net/?f=+%5Cfrac%7B930%2A0.5%2At%2B2100%2A2%2A200%2B3.3%2A10%5E%7B5%7D%2B4200%2A2%2A100%2B2.26%2A10%5E%7B6%7D%2A2++%7D%7B1.3%2A10%5E%7B7%7D+%7D+)
m=
![\frac{465t+6.86*10^{6} }{1.3*10^{7} }](https://tex.z-dn.net/?f=+%5Cfrac%7B465t%2B6.86%2A10%5E%7B6%7D+%7D%7B1.3%2A10%5E%7B7%7D+%7D+)
Я не знаю температуру алюминиевого котелка и гадать не буду. Сверху готовая формула, просто вместо t подставляешь эту температуру и считаешь.
¹⁴⁶₆₂Sm ⇒ ⁴₂He + ᵃₓY
Закон сохранения массового числа:
146=a+4 ,a=142.
Закон сохранения заряда:
62=x+2 .x=60.
ᵃₓY = ¹⁴²₆₀Nd
¹⁴⁶₆₂Sm ⇒ ⁴₂He + ¹⁴²₆₀Nd.
²⁰⁸₈₃Bi ⇒ ⁴₂He + ²⁰⁴₈₁Tl
²¹²₈₄Po ⇒ ⁴₂He + ²⁰⁸₈₂Pb
¹⁴²₅₈Ce ⇒ ⁴₂He + ¹³⁸₅₆Ba
¹⁵¹₆₇Ho ⇒ ⁴₂He + ¹⁴⁷₆₅Tb
¹⁴₆C ⇒ ⁰₋₁e + ¹⁴₇N
1) a = μ1g = 1 м\с в квадрате
2) L = 3\8 (1 - 3μ первое - 2μ второе)gt в квадрате ≈ 75см