Предмет: Физика, автор: sula6108

помогите пожалуйста ​

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Ответы

Автор ответа: andreyvalentev
0

Ответ:

2.Г

3.В

4.В

5.А

6.Г

Объяснение:

хы

Автор ответа: lvv9136
1

Ответ:

Объяснение:

4. V=12 м/c;

Ep=Ek/9;

h   ?     h=Ep/(m*g); Ep=Ek/9=m*V²/18=m*12²/18=8*m; h=8*m/(m*g)=8/10=0,8 м

5. A=25 cм=0,25 м;

A₁=15 cм=0,15 м;

V₁=0,8 м/c;

V max   ?    Повна енергія системи:

Eпmax = k·(Δxmax)²/2 = k·(0,25)²/2 ≈ 0,031·k      Дж    

2) Кінетична енергія: Eк₁ = m·V₁²/2 = m·0,8²/2 = 0,32·m         Дж

Потенційна: Eп₁ = k·(Δx₁)²/2 = k·0,15²/2 ≈ 0,011·k        Дж

Повна енергія: E =Eпmax =Eк₁ + Eп₁ = 0,32·m + 0,011·k   Дж            

0,32·m + 0,011·k = 0,031·k; 0,32·m = 0,020·k; m = 0,020·k / 0,32 = k / 16;

m·(Vmax)²/2 = 0,031·k; k·(Vmax)²/32 = 0,031·k; (Vmax)²= 32·0,031=0,992;

V max = √ (0,992) ≈ 1 м/с Відповідь А

6, Принимая ядра за шарики, забыв о всех квант. эффектах

p₁=m*V(дейтерий 2+1)

p₂=0;

p₁'=-m*V₂=-3*V₂

p₂'=M*V₃=6*V₃.

p₁/p₁'    ?       p₁/p₁'=m*V/m*V₂ ; выделится Е=22,4 MэВ=44,8*10⁻¹³ Дж

Ek₁=3*V²/2=6*V₃²/2+3*V₂²/2. Ед=2,22457 МэВ. Відповідь А

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