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inysia [295]
3 years ago
13

What is the source of the energy released in fission and fusion reactions?

Physics
2 answers:
Alex Ar [27]3 years ago
7 0

Strong nuclear forces

Explanation:

The formation and breakage of the strong nuclear forces in the nucleus of an atom accounts for the energy released during fission and fusion reactions.

Nuclear reactions occurs when changes are made to the nucleus of an atom.

  • Strong nuclear forces maintains the stability in the nucleus of an atom.
  • Strong nuclear forces are the strongest of the fundamental forces.
  • They hold subatomic particles together.
  • When they are broken during a nuclear fission, they release a great deal of energy.
  • When they form during a fusion they also release energy.

learn more;

Nuclear reaction brainly.com/question/4207569

#learnwithBrainly

blagie [28]3 years ago
3 0

Answer:

mass converted to energy

Explanation:

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Find the magnitude and direction of the displacement vector.
Greeley [361]

Answer:

1. The magnitude of the vector is 26.9 m

2. The magnitude of direction of the vector is 21.8°.

Explanation:

From the question given, we obtained the following data:

Displacement along x (Bₓ) = 25 m

Displacement along y (Bᵧ) = 10 m

Magnitude of displacement (B) =?

Magnitude of direction (θ) =.?

1. Determination of the magnitude of displacement of the vector (B)

Displacement along x (Bₓ) = 25 m

Displacement along y (Bᵧ) = 10 m

Magnitude of displacement (B) =?

B² = Bₓ² + Bᵧ²

B² = 25² + 10²

B² = 625 + 100

B² = 725

Take the square root of both side

B = √725

B = 26.9 m

Therefore, the magnitude of the vector is 26.9 m

2. Determination of the magnitude of the direction of the vector.

Displacement along x (Bₓ) = 25 m

Displacement along y (Bᵧ) = 10 m

Magnitude of direction (θ) =.?

The magnitude of direction of the vector can be obtained by using Tan ratio as illustrated below:

Adjacent = Bₓ = 25 m

Opposite = Bᵧ = 10 m

Magnitude of direction (θ) =.?

Tan θ = Opposite /Adjacent

Tan θ = Bᵧ / Bₓ

Tan θ = 10 /25

Tan θ = 0.4

Take the inverse of Tan

θ = Tan¯¹ (0.4)

θ = 21.8°

Therefore, the magnitude of direction of the vector is 21.8°.

7 0
4 years ago
a car slows down from - 27.7 m/s to -10.9 m/s while undergoing a displacement of -105 m .what is its acceleration?
Alja [10]

Answer:

3.09 m/s²

Explanation:

Given:

Δx = -105 m

v₀ = -27.7 m/s

v = -10.9 m/s

Find: a

v² = v₀² + 2aΔx

(-10.9 m/s)² = (-27.7 m/s)² + 2a (-105 m)

a = 3.09 m/s²

4 0
3 years ago
Sayid drew an energy pyramid. What does each section of the triangle represent?
BaLLatris [955]

Pyramid of numbers: relative number of individual organisms at each trophic level in an ecosystem. Pyramids of energy show the relative amount of energy available at each trophic level. On average, about 10 percent of the energy available within one trophic level is transferred to the next trophic level.

HOPE THIS HELPED!!! XD

6 0
3 years ago
Read 2 more answers
1. Using Pj= e^-BE/Z in the Gibb’s form for
Effectus [21]

Answer:

Explanation:

Solution. The Gibbs formula for the entropy is

S = −KB∑pi ㏑(pi).

Using the Boltzmann probability in the canonical ensemble pi = exp(−βEi)/Z, we have,

S = −kB∑ pi[ -(Ei ÷ KBT) - ㏑(Z)] = (U÷T)+ kBln(Z) so U − T S = −kBTln(Z),

where we used∑ P i = 1 , U= ∑i piEi

3 0
3 years ago
The coefficient of linear expansion for some common materials are listed in the table to the right. You would like to design an
Stella [2.4K]

Answer:

A)    T_{f} = 77.6  ⁰C ,  B) T_{f} = 341.6 C

Explanation:

The thermal expansion is given by

           ΔL = α L₀ ΔT

The thermal expansion coefficient for polycarbonate varies between 65 and 70 10⁻⁶ ⁰C⁻¹ and for cast iron 12 10⁻⁶ C⁻¹

Let's look for the final temperature that says same shift change

          ΔT = ΔL / α L₀

          T_{f} - T₀ = ΔL /α L₀

          T_{f}= T₀ + ΔL / α L₀

Let's calculate for each material

A) Polycarbonate use as thermal expansion coefficient 70 10⁻⁶ C⁻¹

         T_{f} = 23.0 + 0.0268 / (70 10⁻⁶ 7.01)

         T_{f} = 23.0 + 54.6

        T_{f} = 77.6  ⁰C

B) cast iron with thermal expansion coefficient 12 10⁻⁶ C⁻¹

      T_{f} = 23.0 + 0.0268 / (12 10⁻⁶ 7.01)

      T_{f} = 23.0 + 3.1859 10²

      T_{f} = 341.6 C

8 0
4 years ago
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