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3 years ago

What is the slit spacing of a diffraction necessary for a 600 nm light to have a first order principal maximum at 25.0°?

Physics

1 answer:

Sauron [17]3 years ago

6 0

Explanation:

Given that,

Wavelength of light, $\lambda=600\ nm=6\times 10^{-7}\ m$

Angle, $\theta=25^{\circ}$

We need to find the slit spacing for diffraction. For a diffraction, the first order principal maximum is given by :

$d\sin\theta=n\lambda$

n is 1 here

d is slit spacing

$d=\dfrac{\lambda}{\sin\theta}\\\\d=\dfrac{6\times 10^{-7}}{\sin(25)}\\\\d=1.41\times 10^{-6}\ m\\\\d=1.41\ \mu m$

So, the slit spacing is $1.41\ \mu m$ .

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The downward pull of an object due to gravity is the objects _______?

elena-s [515]

The downward pull of an object due to gravity is the object’s weight.

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3 years ago

Read 2 more answers

What change in entropy occurs when a 0.15 kg ice cube at -18 °C is transformed into steam at 120 °c 4.

Studentka2010 [4]

Answer: The change in entropy of the given process is 1324.8 J/K

Explanation:

The processes involved in the given problem are:

$1.)H_2O(s)(-18^oC,255K)\rightarrow H_2O(s)(0^oC,273K)\\2.)H_2O(s)(0^oC,273K)\rightarrow H_2O(l)(0^oC,273K)\\3.)H_2O(l)(0^oC,273K)\rightarrow H_2O(l)(100^oC,373K)\\4.)H_2O(l)(100^oC,373K)\rightarrow H_2O(g)(100^oC,373K)\\5.)H_2O(g)(100^oC,373K)\rightarrow H_2O(g)(120^oC,393K)$

Pressure is taken as constant.

To calculate the entropy change for same phase at different temperature, we use the equation:

$\Delta S=m\times C_{p,m}\times \ln (\frac{T_2}{T_1})$ .......(1)

where,

$\Delta S$ = Entropy change

$C_{p,m}$ = specific heat capacity of medium

m = mass of ice = 0.15 kg = 150 g (Conversion factor: 1 kg = 1000 g)

$T_2$ = final temperature

$T_1$ = initial temperature

To calculate the entropy change for different phase at same temperature, we use the equation:

$\Delta S=m\times \frac{\Delta H_{f,v}}{T}$ .......(2)

where,

$\Delta S$ = Entropy change

m = mass of ice

$\Delta H_{f,v}$ = enthalpy of fusion of vaporization

T = temperature of the system

Calculating the entropy change for each process:

For process 1:

We are given:

$m=150g\\C_{p,s}=2.06J/gK\\T_1=255K\\T_2=273K$

Putting values in equation 1, we get:

$\Delta S_1=150g\times 2.06J/g.K\times \ln(\frac{273K}{255K})\\\\\Delta S_1=21.1J/K$

For process 2:

We are given:

$m=150g\\\Delta H_{fusion}=334.16J/g\\T=273K$

Putting values in equation 2, we get:

$\Delta S_2=\frac{150g\times 334.16J/g}{273K}\\\\\Delta S_2=183.6J/K$

For process 3:

We are given:

$m=150g\\C_{p,l}=4.184J/gK\\T_1=273K\\T_2=373K$

Putting values in equation 1, we get:

$\Delta S_3=150g\times 4.184J/g.K\times \ln(\frac{373K}{273K})\\\\\Delta S_3=195.9J/K$

For process 4:

We are given:

$m=150g\\\Delta H_{vaporization}=2259J/g\\T=373K$

Putting values in equation 2, we get:

$\Delta S_2=\frac{150g\times 2259J/g}{373K}\\\\\Delta S_2=908.4J/K$

For process 5:

We are given:

$m=150g\\C_{p,g}=2.02J/gK\\T_1=373K\\T_2=393K$

Putting values in equation 1, we get:

$\Delta S_5=150g\times 2.02J/g.K\times \ln(\frac{393K}{373K})\\\\\Delta S_5=15.8J/K$

Total entropy change for the process = $\Delta S_1+\Delta S_2+\Delta S_3+\Delta S_4+\Delta S_5$

Total entropy change for the process = $[21.1+183.6+195.9+908.4+15.8]J/K=1324.8J/K$

Hence, the change in entropy of the given process is 1324.8 J/K

4 0

3 years ago

Identify which one is chemical change, chemical property,physical change and physical property

Ulleksa [173]

Answer:

Heat conductivity: Physical property

Silver tarnishing: chemical change

Sublimation: physical change

Magnetizing steel: physical property

Length of a metal object: physical property

Shortening melting: physical property

exploding dynamite: chemical change

combustible: chemical property

water freezing: physical change

Acid resistance: chemical property

brittleness: physical property

milk souring: chemical change

baking bread: chemical change

Explanation:

First you need to understand the differences between physical and chemical change; physical and chemical property.

Physical properties vs chemical properties:

Physical properties are properties that you can observe without changing the substance's composition. This means you can measure these properties without changing them chemically.

Chemical properties, on the other hand, are properties are not as direct. These properties are generally determined by the way the react with other substances changing their composition.

Physical changes vs chemical changes:

A substance that undergoes physical change does not change in chemical composition. They may look physically different in terms of size and shape, but overall, their chemical composition remains constant. The best example would be water. Water can change phases, from solid to liquid when they melt. Essentially, they look like different substances, but the change is only physical and not chemical.

Chemical change, is different by the fact that they change in chemical composition. Bonds are broken and/or made through the reaction, which changes them not only physically but chemically as well. Some of the most indicative signs of a chemical change occurring are: change in color, odor, production of gas, production of light/heat.

4 0

3 years ago

The pH of coffee is _______ times greater than the pH of pure water.

DENIUS [597]

Answer:

zero times greater

Explanation:

Coffee's pH level is 5

Pure water's pH level is 7

3 0

3 years ago

Lindsay and coworkers "slime in the first-grade teacher's desk" experiment showed that presenting

tia_tia [17]

The answer that completes the statement is this: A PHOTOGRAPH OF THE PARTICIPANT'S FIRST-GRADE CLASS INCREASED THE LIKELIHOOD OF FALSE MEMORIES. This experiment is similar to Maryanne Gary's experiment which also states that viewing memory as a tape or video recording can be mistaken and may create false memories.

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3 years ago