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Leona [35]
3 years ago
15

What’s the similarity between nuclear fission and nuclear fusion

Physics
1 answer:
Darya [45]3 years ago
4 0

Explanation:

At first sight, it doesn’t make sense that both fission and fusion release energy.

The key is in how tightly the nucleons are held together in a nucleus. If a nuclear reaction produces nuclei that are more tightly bound than the originals, then the excess energy will be released.

It turns out that the most tightly bound atomic nuclei are around the size of iron-56.

Thus, if you split a nucleus that is much larger than iron into smaller fragments, you will release energy because the smaller fragments are at a lower energy than the original nucleus.

If instead you fuse very light nuclei to get bigger products, energy is again released because the nucleons in the products are more tightly bound than in the original nuclei.

https://socratic.org/questions/how-are-fusion-and-fission-similar

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A thin, light wire 75.1 cm long having a circular cross section 0.555 mm in diameter has a 25.4 kg weight attached to it, causin
seraphim [82]

Answer:

(a) 3.23×10⁸ N/m²

(b)  1.46×10⁻³

(c) 2.21×10¹¹ N/m²

Explanation:

(a) Stress = Force/Area.

Stress = F/A................ Equation 1

But,

F = mg................. Equation 2

Where m = mass, and g = acceleration due to gravity

A = πd²/4................. Equation 3

d = diameter of the circular cross section.

Substitute equation 2 and equation 3 into equation 1

Stress = 4mg/πd²............. Equation 4

Given: m = 25.4 kg, d = 0.555 mm = 0.000555 m

Constant: g = 9.8 m/s², π = 3.142

Substitute these values into equation 4

Stress = 4(25.4)(9.8)/(3.142×0.00555²)

Stress = 995.68/(3.08×10⁻⁶)

Stress = 3.23×10⁸ N/m²

(b)

Strain = ΔL/L.............. Equation 5

Where ΔL = extension, L = Length.

Given: ΔL = 1.1 mm = 0.0011 m, L = 75.1 cm = 0.751 m

Substitute into equation 5

Strain = 0.0011/0.751

Strain = 1.46×10⁻³.

(c)

Young modulus = Stress/Strain

Young modulus = 3.23×10⁸/ 1.46×10⁻³

Young modulus = 2.21×10¹¹ N/m²

3 0
3 years ago
A total of 25.6 kJ of heat energy is added to a 5.46 L sample of helium at 0.991 atm. The gas is allowed to expand against a fix
bagirrra123 [75]

Answer:

(a) W = 1329.5 J = 1.33 KJ

(b) ΔU = 24.27 KJ

Explanation:

(a)

Work done by the gas can be found by the following formula:

W = P\Delta V

where,

W = Work = ?

P = constant pressure = (0.991 atm)(\frac{101325\ Pa}{1\ atm}) = 100413 Pa

ΔV = Change in Volume = 18.7 L - 5.46 L = (13.24 L)(\frac{0.001\ m^3}{1\ L}) = 0.01324 m³

Therefore,

W = (100413 Pa)(0.01324 m³)

<u>W = 1329.5 J = 1.33 KJ</u>

<u></u>

(b)

Using the first law of thermodynamics:

ΔU = ΔQ - W (negative W for the work done by the system)

where,

ΔU = change in internal energy of the gas = ?

ΔQ = heat added to the system = 25.6 KJ

Therefore,

ΔU = 25.6 KJ - 1.33 KJ

<u>ΔU = 24.27 KJ</u>

3 0
3 years ago
A sound wave has a frequency of 295 Hz and travels the length of a football field, 91.4 m in 0.506 s. What is the period of the
dem82 [27]

Answer:1/295 seconds

Explanation:

Period=1/frequency

Period=1/295 seconds

7 0
3 years ago
Which of the following should be useful to show the percentage of total growth in plants exposed to various pollutants?
marishachu [46]
I think it is D. Bar graph
5 0
3 years ago
Read 2 more answers
When a potassium atom forms an ion, it loses one electron. What is the electrical charge of the potassium ion? *
Ganezh [65]
+1 An electron has a negative charge so losing a charge of -1 from an uncharged, or neutral, atom will leave an ion with a positive charge.
5 0
3 years ago
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