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3 years ago

What happens to the diffraction pattern when the number of lines per centimeter of a diffraction grating is increased?

1 answer:

3 0

The separation between two neighboring vital maxima increments. As the quantity lines per centimeter expand, the detachment d between neighboring openings ends up noticeably littler. The point that determines primary maxima of a diffraction grinding is given by transgression = m/d, where is the wavelength and m = 0, 1, 2, 3. For settled estimations of m and the wavelength, this connection demonstrates that as d abatements, increments. For a settled screen area, this implies the separation between two neighboring chief maxima additionally increments.

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To what volume will a sample of gas expand if it is heated from 50.0°c and 2.33 l to 500.0°c?

Lena [83]

The volume of the gas at $500^{\circ} \text{C}$ is $\fbox{\begin\ 5.576\,{\text{L}}\\\end{minispace}}$ .

Further Explanation:

Consider the pressure of the gas to be constant.

The change in the volume of the gas when the temperature of the gas is varied by keeping the pressure of the gas at a constant value is defined by the Charles' Law.

Concept:

According to the Charles law, the volume of the gas is directly proportional to the temperature of the gas at constant pressure.

The Charles' law can be stated as:

$\fbox{\begin\\V\propto T\\\end{minispace}}$

The above expression can we written as.

$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

Convert the temperature of the gas into kelvin.

$T=273+T^\circ\text{C}}}$

Here, T is the temperature in kelvin and $T^\circ\text{C}}}$ is the temperature in centigrade.

The initial temperature of the gas is $50^\circ\text{C}$ . The temperature of the gas in kelvin is.

$\begin{aligned}{T_1}&=273+50\\&=323\,{\text{K}}\\\end{aligned}$

The final temperature of the gas is $500^\circ\text{C}$ . The temperature in kelvin is.

$\begin{aligned}{T_2}&=273+500\\&=773\,{\text{K}}\\\end{aligned}$

Substitute the values of temperature and volume in the expression of the Charles' Law.

$\begin{aligned}{V_2}&=\frac{{{T_2}}}{{{T_1}}}{V_1}\\&=\frac{{773\,{\text{K}}}}{{323\,{\text{K}}}}\left({2.33\,{\text{L}}}\right)\\&=5.576\,{\text{L}}\\\end{aligned}$

Thus, the volume of the gas at $500^\circ\text{C}}$ will be $\fbox{\begin\ 5.576\,{\text{L}}\\\end{minispace}}$ .

Learn More:

1. Examples of wind and solar energy brainly.com/question/1062501

2. Stress developed in a wire brainly.com/question/12985068

Answer Details:

Grade: High school

Subject: Physics

Chapter: Gas law

Keywords:

Charles law, temperature, volume, initial, final, kelvin, centigrade, 50 C, 500 degree, 500 C, 50 degree, 2,33 L, gas expand, sample, heated.

7 0

3 years ago

Determine the average value of the translational kinetic energy of the molecules of an ideal gas at (a) 27.8°C and (b) 143°C. Wh

Alinara [238K]

Answer:

a) $k_{avg}=6.22\times 10^{-21}$

b) $k_{avg}=8.61\times 10^{-21}$

c) $k_{mol}=3.74\times 10^{3}J/mol$

d) $k_{mol}=5.1\times 10^{3}J/mol$

Explanation:

Average translation kinetic energy ( $k_{avg}$ ) is given as

$k_{avg}=\frac{3}{2}\times kT$ ....................(1)

where,

k = Boltzmann's constant ; 1.38 × 10⁻²³ J/K

T = Temperature in kelvin

a) at T = 27.8° C

T = 27.8 + 273 = 300.8 K

substituting the value of temperature in the equation (1)

we have

$k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 300.8$

$k_{avg}=6.22\times 10^{-21}J$

b) at T = 143° C

T = 143 + 273 = 416 K

substituting the value of temperature in the equation (1)

we have

$k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 416$

$k_{avg}=8.61\times 10^{-21}J$

c ) The translational kinetic energy per mole of an ideal gas is given as:

$k_{mol}=A_{v}\times k_{avg}$

here $A_{v}$ = Avagadro's number; ( 6.02×10²³ )

now at T = 27.8° C

$k_{mol}=6.02\times 10^{23}\times 6.22\times 10^{-21}$

$k_{mol}=3.74\times 10^{3}J/mol$

d) now at T = 143° C

$k_{mol}=6.02\times 10^{23}\times 8.61\times 10^{-21}$

$k_{mol}=5.1\times 10^{3}J/mol$

8 0

3 years ago

Big Ben, a large artifact in England, has a mass of 1x10^8 kilograms and the Empire State Building 1x10^9 kilograms. The distanc

TiliK225 [7]

Answer:

The force, exerted by Big Ben on the Empire State Building is 2.66972 × 10⁻⁷ N

Explanation:

The question relates to the force of gravity experienced between two bodies

The given parameters are;

The mass of Big Ben, M₁ = 1 × 10⁸ kg

The mass of the Empire State Building, M₂ = 1 × 10⁹ kg

The distance between the two Big Ben and the Empire State Building, r = 5,000,000 meters

By Newton's Law of gravitation, we have;

$F=G \times \dfrac{M_{1} \times M_{2}}{r^{2}}$

Where;

F = The force exerted by Big Ben on the Empire State Building and vice versa

G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

M₁, M₂, and r are the given parameters

By plugging in the values of the parameters and the constant into the equation for Newton's Law of gravitation, we have;

$F=6.67430 \times 10^{-11} \times \dfrac{1 \times 10^8 \times 1 \times 10^9}{(5,000,000)^{2}} = 2.66972 \times 10^{-7}$

The force, 'F', exerted by Big Ben on the Empire State Building is F = 2.66972 × 10⁻⁷ N.

3 0

3 years ago

A 5.0-kg block of wood is placed on a 2.0-kg aluminum frying pan. How much heat is required to raise the temperature of both the

Shalnov [3]

Heat required to raise the temperature of a given system is

$Q = ms\Delta T$

here we know that

m = mass

s = specific heat capacity

$\Delta T$ = change in temperature

now as we know that

mass of wood = 5 kg

mass of aluminium pan = 2 kg

change in temperature = 45 - 20 = 25 degree C

specific heat capacity of wood = 1700 J/kg C

specific heat capacity of aluminium = 900 J/kg C

now here we will find the total heat to raise the temperature of both

$Q = m_1s_1\Delta T_1 + m_2s_2\Delta T_2$

$Q = 5 * 1700 * 25 + 2 * 900 * 25$

$Q = 212500 + 45000$

$Q = 257500 J$

So heat required to raise the temperature of the system is 257500 J

4 0

3 years ago

Which types of numbers does scientific notation best describe?

Travka [436]

The correct answer is
c) very small and very large

Let's see this with a few examples:
1) if we have a very small number, such as
 $0.0000000001$
we see that we can write it easily by using the scientific notation:
 $1\cdot 10^{-10}$
2) Similarly, if we have a very large number:
 $10000000000$
we see that we can write it easily by using again the scientific notation:
 $1 \cdot 10^{10}$

4 0

3 years ago