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shtirl [24]
3 years ago
15

A star has a surface temperature of 30,000 K. At what wavelength does the peak radiation occur? 100 nm

Physics
2 answers:
tangare [24]3 years ago
8 0
Use the displacement law, peak wavelength = 0.0029/T =0.0029/30000 = 97nm
Eva8 [605]3 years ago
8 0

Answer : Wavelength is 97 nm.

Explanation :

It is given that,

The surface temperature of a star, T=30,000\ K

We have to find the wavelength where the peak radiation occurs.

We can find the wavelength using Wein's displacement law i.e.

\lambda=\dfrac{b}{T}

b is the Wein's constant b=2.88\times 10^{-3}m.K

So, \lambda=\dfrac{2.88\times 10^{-3}\ m.K}{30,000\ K}

\lambda=9.6\times 10^{-8}\ m

or \lambda=97\ nm

Hence, the correct option is (b).        

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a circular cylinder and isused to maintain a water depth of 4 m. That is, when the water depth exceeds 4 m, thegate opens slight
stiv31 [10]

Answer:

  W / A = 39200 kg / m²

Explanation:

For this problem let's use the equilibrium equation of / newton

           F = W

Where F is the force of the door and W the weight of water

         W = mg

We use the concept of density

        ρ = m / V

        m = ρ V

The volume of the water column is

          V = A h

We replace

         W = ρ A h g

On the other side the cylinder cover has a pressure

          P = F / A

          F = P A

We match the two equations

       P A = ρ A h g

        P = ρ g h

        P = 39200 Pa

The weight of the water column is

       W  = 1000 9.8 4 A

       W / A = 39200 kg / m²

3 0
3 years ago
A diffraction grating with 750 slits per mm is illuminated by light which gives a first-order diffraction angle of 34.0°. What i
lesya [120]
<h2>Answer: 745.59 nm</h2>

Explanation:

The diffraction angles \theta_{n} when we have a slit divided into n parts are obtained by the following equation:

dsin\theta_{n}=n\lambda (1)

Where:

d is the width of the slit

\lambda is the wavelength of the light  

n is an integer different from zero

Now, the first-order diffraction angle is given when n=1, hence equation (1) becomes:

dsin\theta_{1}=\lambda (2)

We know:

\theta_{1}=34\°

In addition we are told the diffraction grating has 750 slits per mm, this means:

d=\frac{1mm}{750}

Solving (2) with the known values we will find \lambda:

\lambda=(\frac{1mm}{750})sin(34\°) (3)

\lambda=0.00074559mm (4)

Knowing 1mm=10^{6}nm:

\lambda=745.59nm  >>>This is the wavelength of the light, wich corresponds to red.

6 0
4 years ago
For any molecule, formula unit, or ion, the sum of the average atomic masses of all the atoms represented in a formula is the.
alukav5142 [94]

The mass of a substance is given in atomic mass units and is calculated by adding the average atomic masses of all the atoms in the substance's chemical formula.

<h3>What empirical formula represents the total average atomic mass of every atom?</h3>

The Method The average atomic masses of all the atoms included in a formula's representation are added to get the mass of any molecule, formula unit, or ion. It has no bearing on the number of significant figures because the number of atoms is an exact quantity. One H2O molecule weighs 18.02 amu on average.

<h3>What connection exists between the empirical formula and the molecular formula?</h3>

You can determine the number of atoms of each element in a molecule using its molecular formula. These empirical formulations provide the most basic or reduced elemental ratio of a compound. The empirical formula and the molecular formula of a substance are same if the molecular formula can no longer be decreased.

To know more about atomic mass visit:-

brainly.com/question/17067547

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5 0
1 year ago
A thin glass rod is submerged in oil. What is the critical angle for light traveling inside the rod? The index of refraction for
Alekssandra [29.7K]
When light travels from a medium with higher refractive index to a medium with lower refractive index, the critical angle is the angle of incidence above which light is reflected only (no refraction occurs), and the value of this critical angle is given by
\theta_c = \arcsin ( \frac{n_2}{n_1} )
where n2 is the refractive index of the second medium and n1 is the refractive index of the first medium.

In this problem, the first medium is the glass (n_1 = 1.50), while the second medium is oil (n_2 =1.46), therefore the critical angle is given by
\theta_c = \arcsin( \frac{1.46}{1.50} )=\arcsin(0.973)=76.7^{\circ}
7 0
3 years ago
If the pressure inside the cylinder increases to 1.6 atm, what is the final volume, in milliliters, of the cylinder?
Bess [88]

This is an incomplete question, here is a complete question.

The air in the cylinder with a piston has a volume of 220 mL and a pressure of 650 mmHg.

If the pressure inside the cylinder increases to 1.6 atm, what is the final volume, in milliliters, of the cylinder?

Answer : The final volume of the cylinder is, 117.6 mL

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}

or,

P_1V_1=P_2V_2

where,

P_1 = initial pressure of gas = 650 mmHg = 0.855 atm   (1 atm = 760 mmHg)

P_2 = final pressure of gas = 1.6 atm

V_1 = initial volume of gas = 220 mL

V_2 = final volume of gas = ?

Now put all the given values in the above equation, we get:

0.855atm\times 220mL=1.6atm\times V_2

V_2=117.6mL

Therefore, the final volume of the cylinder is, 117.6 mL

8 0
3 years ago
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