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Thepotemich [5.8K]
3 years ago
13

The wavelength of yellow sodium light in vacuum is approximately 6.0 x 10-7m. The frequency

Physics
1 answer:
Ludmilka [50]3 years ago
5 0

The frequency of sodium light in vacuum is 5 ×10^{14} Hz.

Answer:

Option e

Explanation:

It is known that frequency of a wave is inversely proportional to the wavelength of the wave. And the proportionality constant is the speed of light in vacuum. As the speed of light in vacuum is known as 3×10^{8} m/s and the wavelength of the light in vacuum is given as 6 × 10^{-7} m, then the frequency of the light is determined as

Frequency = Speed of light in vacuum/Wavelength of light

Frequency = \frac{3*10^{8} }{6*10^{-7} }

Frequency = 0.5×10^{15} = 5 ×10^{14} Hz.

So the frequency of sodium light in vacuum is 5 ×10^{14} Hz.

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The wheel of a stationary exercise bicycle at your gym makes one rotation in 0.670 s. Consider two points on this wheel: Point P
lesya [120]

Answer:

a) For P: v=0.938\frac{m}{s}

For Q: v = 1.876\frac{m}{s}

b) For P:

a_{rad}=8.80\frac{m}{s^{2}}

for Q:

a_{rad}=17.60\frac{m}{s^{2}}

c) As the distance from the axis increases then speed increases too.

Explanation:

a) Assuming constant angular acceleration we can find the angular speed of the wheel dividing the angular displacement θ between time of rotation:

\omega =\frac{\theta}{t}

One rotation is 360 degrees or 2π radians, so θ=2π

\omega =\frac{2\pi}{0.670} =9.38\frac{rad}{s}

Angular acceleration is at every point on the wheel, but speed (tangential speed) is different and depends on the position (R) respect the rotation axis, the equation that relates angular speed and speed is:

v = \omega R

for P:

v = 9.38\frac{rad}{s}*0.1m=0.938\frac{m}{s}

for Q:

v = 9.38\frac{rad}{s}*0.2m=1.876\frac{m}{s}

b) Centripetal acceleration is:

a_{rad}= \frac{v^2}{R}

for P:

a_{rad}= \frac{(0.938)^2}{0.1}=8.80\frac{m}{s^{2}}

for Q:

a_{rad}= \frac{(1.876)^2}{0.2}=17.60\frac{m}{s^{2}}

c) As seen on a) speed and distance from axis is v = \omega R because ω is constant the if R increases then v increases too.

3 0
3 years ago
How many electrons will a single atom of sulfur with no charge and no bonds have in its valence shell?
Svet_ta [14]

Answer:

There are 6 electrons in the outermost shell.

Explanation:

Sulphur is a non-mettalic element which is in the period 3 and group .6on the periodic table. It has an atomic number of 16 and a Mass number of 32. Atomic number tells you the number of electrons in an electrically neutral atom. It has the electronic configuration of 1s2 2s2 2p6 3s2 3p4.

The orbitals have a formula 2n^2 where n = 0, 1, 2, 3 etc.

In the shells, n = 1 so there are 2 electrons. For n = 2, 2*(2)^2 = 8 electrons. So, 16 - (8 + 2) = 6 electrons in the 3 shell (outermost shell)

Therefore from the electronic confriguration above, there are 6 electrons in the outermost shell.

6 0
3 years ago
Read 2 more answers
A diffraction grating is placed 1.00 m from a viewing screen. Light from a hydrogen lamp goes through the grating. A hydrogen sp
Colt1911 [192]

Answer:

λ = 396.7 nm

Explanation:

For this exercise we use the diffraction ratio of a grating

           d sin θ = m λ

in general the networks works in the first order m = 1

we can use trigonometry, remembering that in diffraction experiments the angles are small

           tan θ = y / L

           tan θ = \frac{sin \theta}{cos \theta} = sin θ

           sin θ = y / L

we substitute

          d \  \frac{y}{L} = m λ

with the initial data we look for the distance between the lines

           d = \frac{m \lambda \ L}{y}

           d = 1 656 10⁻⁹ 1.00 / 0.600

            d = 1.09 10⁻⁶ m

for the unknown lamp we look for the wavelength

           λ = d y / L m

           λ = 1.09 10⁻⁶ 0.364 / 1.00 1

           λ = 3.9676 10⁻⁷ m

           λ = 3.967 10⁻⁷ m

         

we reduce nm

           λ = 396.7 nm

8 0
3 years ago
A constant voltage of 12.00 V has been observed over a certain time interval across a 1.20 H inductor. The current through the i
kiruha [24]

Answer:

The time is 0.5 sec.

Explanation:

Given that,

Voltage V= 12.00 V

Inductance L= 1.20 H

Current = 3.00 A

Increases rate = 8.00 A

We need to calculate change in current

\Delta A = 8.00-3.00= 5.00\ A

We need to calculate the time interval

Using formula of inductor

V=L\dfrac{\Delta A}{\Delta t}

\Delta t =\dfrac{L\Delta A}{V}

Where, \Delta A = change in current

V = voltage

L = inductance

Put the value into the formula

\Delta t=\dfrac{1.20\times5.00}{12.00}

\Delta t=0.5\ sec

Hence, The time is 0.5 sec.

5 0
3 years ago
The significant feature of a Cepheid variable is that there is a relationship between two intrinsic parameters, one of which can
Ghella [55]

Answer:

Period of brightness variation and luminosity.

Explanation:

The Cepheid variables are used as distance indicators. This requires estimation of periods and (usually) intensity-mean magnitudes in order to establish a period—apparent luminosity relation. It is particularly important for the techniques employed to be as accurate and efficient as possible.

5 0
3 years ago
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