There are strong winds on these trees. What makes the wind stronger? Why are they blowing from different directions?

Calculate the wavelengths of the first five members of the Lyman series of spectral lines, providing the result in units Angstro

Oduvanchick [21]

Answer:

Explanation:

The formula for hydrogen atomic spectrum is as follows

energy of photon due to transition from higher orbit n₂ to n₁

$E=13.6 (\frac{1}{n_1^2 } - \frac{1}{n_2^2})eV$

For layman series n₁ = 1 and n₂ = 2 , 3 , 4 , ... etc

energy of first line

$E_1=13.6 (\frac{1}{1^2 } - \frac{1}{2 ^2})$

10.2 eV

wavelength of photon = 12375 / 10.2 = 1213.2 A

energy of 2 nd line

$E_2=13.6 (\frac{1}{1^2 } - \frac{1}{3 ^2})$

= 12.08 eV

wavelength of photon = 12375 / 12.08 = 1024.4 A

energy of third line

$E_3=13.6 (\frac{1}{1^2 } - \frac{1}{4 ^2})$

12.75 e V

wavelength of photon = 12375 / 12.75 = 970.6 A

energy of fourth line

$E_4=13.6 (\frac{1}{1^2 } - \frac{1}{5 ^2})$

= 13.056 eV

wavelength of photon = 12375 / 13.05 = 948.3 A

energy of fifth line

$E_5=13.6 (\frac{1}{1^2 } - \frac{1}{6 ^2})$

13.22 eV

wavelength of photon = 12375 / 13.22 = 936.1 A

7 0

3 years ago

Violet light (λ = 400 nm) passing through a diffraction grating for which the slit spacing is 6.0 μm forms a pattern on a screen

sergey [27]

Answer:

3 order dark fringe

Explanation:

y = Distance from central bright fringe = 204 mm

λ = Wavelength = 400 nm

L = Distance between screen and source = 1 m

d = Slit distance = 6 μm

$tan\theta =\frac{y}{L}\\\Rightarrow tan\theta =\frac{204}{1000}=0.2012^{\circ}$

$dsin\theta=m\lambda\\\Rightarrow m=\frac{dsin\theta}{\lambda}\\\Rightarrow m=\frac{6\times 10^{-6}sin0.2012}{400\times 10^{-9}}=2.9982\approx 3$

Order of fringe is 3

So, it is a Dark order fringe

3 0

3 years ago

The force of gravity typically acts in a downward direction on the object on earth when an object is standing still, on the flat

Alja [10]

The normal force acts to counter the gravitational force, that is the upward direction.

5 0

3 years ago

An aluminum-alloy rod has a length of 10.0 cm at 20°C and a length of 10.015 cm at the boiling point of water (1000C). (a) What

nikitadnepr [17]

Answer:

a. 9.99625 cm b. 68 °C

Explanation:

(a) What is the length of the rod at the freezing point of water (0 0C)?

Before we find the length of the rod, we need to find the coefficient of linear expansion, α = (L - L₀)/[L₀(T - T₀)] where L₀ = length of rod at temperature T₀ = 10.0 cm, T₀ = 20 °C, L = length of rod at temperature T = 10.015 cm and T = 100 °C

Substituting the values of the variables into the equation, we have

α = (L - L₀)/[L₀(T - T₀)]

α = (10.015 cm - 10.0 cm)/[10.0 cm(100 °C - 20 °C)]

α = 0.015 cm/[10.0 cm × 80 °C]

α = 0.015 cm/[800.0 cm °C]

α = 0.00001875 /°C

We now find the length L₁ at T₁ = 0 °C from

L₁ = L₀(1 + α(T₁ - T₀))

So, substituting the values of the variables into the equation, we have

L₁ = L₀(1 + α(T₁ - T₀))

L₁ = 10.0 cm[1 + 0.00001875 /°C(0° C - 20 °C)]

L₁ = 10.0 cm[1 + 0.00001875 /°C × -20° C]

L₁ = 10.0 cm[1 - 0.000375]

L₁ = 10.0 cm[0.999625]

L₁ = 9.99625 cm

(b) What is the temperature if the length of the rod is 10.009 cm?

With length L₃ = 10.009 cm at temperature T₃, using

L₃ = L₀(1 + α(T₃ - T₀))

making T₃ subject of the formula, we have

L₃/L₀ = 1 + α(T₃ - T₀)

L₃/L₀ - 1 = α(T₃ - T₀)

T₃ - T₀ = (L₃/L₀ - 1)/α

T₃ = T₀ + (L₃/L₀ - 1)/α

substituting the values of the variables into the equation, we have

T₃ = 20 °C + (10.009 cm/10.0 cm - 1)/0.00001875 /°C

T₃ = 20 °C + (1.0009 - 1)/0.00001875 /°C

T₃ = 20 °C + 0.0009/0.00001875 /°C

T₃ = 20 °C + 48 °C

T₃ = 68 °C

8 0

3 years ago

Riley is cleaning his room(a once a year activity) and pulls the Hoover with a force of 8 N a total distance of 20 metres. How m

Radda [10]

Answer:

160J

Explanation:

Given force = 8N and total distance = 20 meters

Workdone = force x distance

= 8 x 20

= 160J

Therefore, workdone by Riley in pulling the hoover is 160J

7 0

3 years ago