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MariettaO [177]
2 years ago
13

Review. Consider the deuterium-tritium fusion reaction with the tritium nucleus at rest:

Physics
1 answer:
asambeis [7]2 years ago
7 0

Considering the deuterium-tritium fusion reaction with the tritium nucleus at rest: ¹₂H + ¹₃H → ²₄He + ⁰₁n  the electric potential energy (in electron volts) at this distance is 17.58MeV

<h3>How is the electric potential energy of deuterium-tritium fusion reaction calculated?</h3>

The reaction is  ¹₂H + 1₃H → ²₄He + ⁰₁n

Value of Q = (Mass of ¹₂H + Mass of ¹₃H - Mass of ²₄He- Mass of n) x 931 MeV

Mass of ¹₂H = 2.014102

Mass of ¹₃H = 3.016049

Mass of ²₄He = 4.002603

Mass of n = 1.00867

Therefore Value of Q = [2.014102+3.016049−4.002603−1.00867] × 931 MeV

Therefore Value of Q = 0.01887 × 931 MeV

= 17.58MeV

To learn more about  deuterium-tritium fusion reaction, refer

brainly.com/question/9054784

#SPJ4

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A railroad freight car rolls on a track at 3.50 m/s toward two identical coupled freight cars, which are rolling in the same dir
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Answer:

Explanation:

Hello! To solve this problem we must be clear about the concept of energy conservation, and kinetic energy with the following sentence

The kinetic energy of the two cars (v = 1.2m / S) plus the kinetic energy of the third car (v = 3.5m / S) must be equal to the kinetic energy of the three cars together.

The kinetic energy is calculated by the following equation.

E=0.5mV^2

m= mass of the cars=26500kg

V=speed

E=kinetic energy

taking into account the above, the following equation is inferred

1=  the cars are separated

2= the cars are togheter

E1=E2

E1=0.5mV1^2+0.5mV1^2+0.5m(Va)^2

where

m= mass of each car

V1= 1.2m/s

Va=3.5,m/S

E2=0.5(3)(m)V^2

m= mass of each car

V=speed (in m/s) of the three coupled cars after the first couples with the other two

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0.5mV1^2+0.5mV1^2+0.5m(Va)^2=0.5(3)(m)V^2

V1^2+V1^2+(Va)^2=(3)V^2.\\2V1^2+(Va)^2=(3)V^2\\V^2=\frac{2V1^2+(Va)^2}{3} \\

V=\sqrt{\frac{2V1^2+(Va)^2}{3}} \\V=\sqrt{\frac{2(1.2)^2+(3.5)^2}{3}} \\\\V=2.245m/s

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7 0
4 years ago
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Two forces act on a 41-kg object. One force has magnitude 65 N directed 59° clockwise from the positive x-axis, and the other ha
Genrish500 [490]

Answer:

a = 1.41 m/s²

Explanation:

Given that

mass ,m= 41 kg

F₁ = 65 N , θ = 59°

F₂ = 35 N ,θ = 32°

The component of Force F₁

F₁x= F₁cos59° i

F₁x= 65 x cos59° i = 33.47 i

F₁y= - F₁ sin 59° j

F₁y= - 65 x sin 59° j = - 55.71 j

The component of Force F₂

F₂x= F₂ sin 32° i

F₂x= 35 x sin 32° i = 18.54 i

F₂y=  F₂ cos 32° j

F₂y=  35 x cos 32° j =  29.68 j

The total force F

F= 33.47 i +   18.54 i - 55.71 j +   29.68 j

F= 52.01 i - 26.03 j

The magnitude of the force F

F=\sqrt{52.01^2+26.03 ^2}\ N

F=58.16 N

We know that

F= m a

a= Acceleration

m=mass

58.16 = 41 x a

a = 1.41 m/s²

6 0
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