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loris [4]
3 years ago
6

The term big bang nucleosynthesis refers to the formation during the big

Physics
1 answer:
Reil [10]3 years ago
3 0

Answer:

D. Light elements

Explanation:

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3. A cat walks 0.220km North, then 0. 120 km South in a time of 400 seconds. whats the displacement and average velocity?
Andru [333]

Answer:

The rate at which velocity changes with respect to a change in time is called. acceleration.

Explanation:

4 0
3 years ago
What occurs when an unstable atomic nucleus changes into other nuclei by emitting particles and energy
Klio2033 [76]
I'm not too sure but I think it's nuclear decay
6 0
3 years ago
A heat engine with 0.300 mol of a monatomic ideal gas initially fills a 1000 cm3 cylinder at 500 K . The gas goes through the fo
LuckyWell [14K]

Complete Question:

A heat engine with 0.300 mol of a monatomic ideal gas initially fills a 1000 cm3 cylinder at 500 K . The gas goes through the following closed cycle: - Isothermal expansion to 5000 cm3. - Isochoric cooling to 400 K . - Isothermal compression to 1000 cm3. - Isochoric heating to 500 K .

a) what is the work for one cycle

b) what is the thermal efficiency

Answer:

a) Work done for 1 cycle = 402.13

b) Thermal efficiency = 20%

Explanation:

Number of moles, n = 0.300 mol

Initial Volume, V₁ = 1000 cm³

Temperature, T = 500 K

Isothermal expansion to 5000 cm³

Final volume, V₂ = 5000 cm³

R = 8.314 J/ mol.K

Work done, W = nRT ln(V₂/V₁)

W = (0.3 * 8.314 * 500) * ln(5000/1000)

W = 1247.1 * ln5

W₁ = 2007.13 J

Isochoric cooling

In an Isochoric process, volume is constant i.e. V₂ = V₁ = V

W = nRT ln(V/V)

But  ln(V/V) = ln 1 = 0

Work done, W₂ = 0 Joules

Isothermal Compression to 1000 cm³

V₂ = 1000 cm³

V₁ = 5000 cm³

W = nRT ln(V₂/V₁)

W = 0.3 * 8.314 * 400 ln(1000/5000)

W₃ = -1605 J

Isochoric heating to 500 K

Since there is no change in volume, no work is done

W₄ = 0 J

a) Work done for 1 cycle

W = W₁ + W₂ + W₃ + W₄

W = 2007.13 + 0 + 0 -1605+0

W = 402.13 Joules

b) Thermal efficiency

Thermal efficiency = (Net workdone for 1 cycle)/(Heat absorbed)

Heat absorbed = Work done due to thermal expansion = 2007.13 J

Thermal efficiency = 402.13/2007.13

Thermal efficiency = 0.2

Thermal efficiency = 0.2 * 100% = 20 %

3 0
3 years ago
Which is an example of transforming potential energy to kinetic energy? Select two options.
Rasek [7]

<u>Complete Question:</u>

Which is an example of transforming potential energy to kinetic energy? Select two options.

changing thermal energy to electrical energy

changing chemical energy to thermal energy

changing nuclear energy to radiant energy

changing radiant energy to electrical energy

changing mechanical energy to chemical energy

<u>Correct Answer:</u>

The examples of transforming potential energy to kinetic energy are changing chemical energy to thermal energy and changing nuclear energy to radiant energy.

<u>Explanation:</u>

As stated by the conservation of energy law, any form of energy is usually transferred to another form. The basic kinds of energy is potential and kinetic energy. Potential energy is the energy stored for the objects at rest and kinetic energy is the energy utilized by the objects for motion.

So in the given options, chemical energy is the energy stored by the chemical bonds to make a stable compound and that energy is converted to thermal energy when the bonds get broken. So the stored energy or the energy required to keep the bonds intact is chemical energy and it is thus a form of potential energy.

And when these bonds get broken, the electrons use the thermal energy released by this breakage as their kinetic energy. So one form of transforming potential energy to kinetic energy is by changing chemical energy to thermal energy.

Similarly, the nuclear energy is exhibited by the elementary particles in an atom. So it is similar to potential energy and the radiant energy is released whenever there is an excitation. So the radiant energy will be similar to kinetic energy.

Thus, the changing of chemical energy to thermal energy and the changing of nuclear energy to radiant energy are the examples of transforming potential energy to kinetic energy.

5 0
3 years ago
A slit of width 2.0 μm is used in a single slit experiment with light of wavelength 650 nm. If the intensity at the central maxi
Valentin [98]

Answer:

The intensity at 10° from the center is 3.06 × 10⁻⁴I₀

Explanation:

The intensity of light I = I₀(sinα/α)² where α = πasinθ/λ

I₀ = maximum intensity of light

a = slit width = 2.0 μm = 2.0 × 10⁻⁶ m

θ = angle at intensity point = 10°

λ = wavelength of light = 650 nm = 650 × 10⁻⁹ m

α = πasinθ/λ

= π(2.0 × 10⁻⁶ m)sin10°/650 × 10⁻⁹ m

= 1.0911/650 × 10³

= 0.001679 × 10³

= 1.679

Now, the intensity I is

I = I₀(sinα/α)²

= I₀(sin1.679/1.679)²

= I₀(0.0293/1.679)²

= 0.0175²I₀

= 0.0003063I₀

= 3.06 × 10⁻⁴I₀

So, the intensity at 10° from the center is 3.06 × 10⁻⁴I₀

5 0
3 years ago
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