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zimovet [89]
3 years ago
11

An electric current through neon gas produces several distinct wavelengths of visible light. What are the wavelengths (in nm) of

the neon spectrum, if they form first-order maxima at angles of 49.67°, 50.65°, 52.06°, and 52.89° when projected on a diffraction grating having 11,000 lines per centimeter? (Round your answers to the nearest nanometer. Due to the nature of this problem, do not use rounded intermediate values in your calculations—including answers submitted in WebAssign. Enter your answers from smallest to largest.)
Physics
1 answer:
MA_775_DIABLO [31]3 years ago
4 0

Answer:

Explanation:

If a be grating element

a = 1 x 10⁻² / 11000

= .0909 x 10⁻⁵

= 909 x 10⁻⁹ m

for first order maxima , the condition is

a sinθ = λ where λ is wavelength

909 x 10⁻⁹ sin 49.67 = λ₁

λ₁ = 692.95  nm .

λ₂ = 909 x 10⁻⁹ sin 50.65

= 702.91 nm

λ₃ = 909 x 10⁻⁹ sin 52.06

= 716.88 nm

λ₄ = 909 x 10⁻⁹ sin 52.89

= 724.90 nm

692.95  nm ,  702.91 nm  , 716.88 nm , 724.90 nm .

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Vector C has a magnitude of 24.6 m and points in the − y ‑ direction. Vectors A and B both have positive y ‑ components, and mak
frez [133]

Answer:

A= 61.35

B= -44.40

Explanation:

1. Using the components method we have:

A_{x}= A cos \alpha\\B_{x}= B cos \beta\\C_{x}= 0\\\\A_{y}= A sin \alpha\\B_{y}= B sin \beta\\C_{y}= 24.6\\

Considering that the vector sum A+B+C=0, then:

|V|=\sqrt{V_{x}^{2} +V_{y}^{2} }=0

Then:

V_{x} ^{2} =0; V_{x} =0\\V_{y} ^{2} =0; V_{y} =0

It means the value of x and y component is 0.

2. Determinate the equations that describe each component:

V_{x}= A cos \alpha -B cos \beta=0  (1)\\V_{y}= A sin \alpha +B sin \beta - C=0   (2)

Form Eq. (1):

A=B \frac{cos \beta}{cos \alpha}     (3)

Replacing A in Eq. (2):

(B\frac{cos \beta}{cos \alpha})(sin \alpha)+ B sin\beta-C=0\\(B\frac{cos \beta}{cos \alpha})(sin \alpha)+ B sin\beta-=C\\\\B(\frac{cos \beta. sin \alpha}{cos \alpha}+ sin\beta)=C\\B=C(\frac{cos \beta. sin \alpha}{cos \alpha}+ sin\beta)^{-1}     (4)

Replacing values of C, α and β in (4):

B= 24.6 (\frac{(cos 27.7)(sin 44.9)}{cos 44.9}+sin 27.7)^{-1}  \\B= -44.4

Replacing value of B in (3)

A=-44.40\frac{cos 27.7}{sin 49.9} \\A= 61.35

5 0
3 years ago
An observer is approaching at stationary source at 17.0 m/s. Assuming the speed of sound is 343 m/s, what is the frequency heard
UNO [17]

Answer:

the frequency heard by the observer is equal to 2677 Hz

Explanation:

given,      

velocity of the observer = 17 m/s

speed of the sound = 343 m/s    

velocity of the source = 0 m/s    

frequency emitted from the source  = 2550 Hz              

f = f_0(\dfrac{v-v_0}{v-v_s})              

f = 2550\times (\dfrac{343+17}{343-0})

velocity of observer is negative as it is approaching the source.                   f = 2676.38 Hz ≈ 2677 Hz                    

hence, the frequency heard by the observer is equal to 2677 Hz

7 0
3 years ago
An 82 kg man, at rest, drops from a diving board 3.0 m above the surface of the water and comes to rest 0.55 s after reaching th
ddd [48]

Answer:

1626.4 N

Explanation:

Given that a 82 kg man, at rest, drops from a diving board 3.0 m above the surface of the water and comes to rest 0.55 s after reaching the water. What force does the water exert on him?

The parameters to be considered are:

Distance S = 3m

Time t = 0.55s

Since the man started from rest, initial velocity u = 0

Using second equation of motion

S = Ut + 1/2at^2

3 = 1/2 × a × 0.55^2

3 = 1/2 × a × 0.3025

a = 3/ 0.15125

a = 19.83 m/s^2

Force = mass × acceleration

Force = 82 × 19.83

Force = 1626.4 N

Therefore, the force that water exerted on him is 1626.4 N

4 0
2 years ago
A force of 880 newtons stretches a spring 4 meters. A mass of 55 kilograms is attached to the end of the spring and is initially
zysi [14]

Answer:

x(t) = -8sin2t

Explanation:

See the attachment for solution

From my solving, we can deduce that w² = 4, and thus, w = 2

Therefore, the general solution is

x(t) = c1 cos2t + c2 sin2t

Considering the final variable, we can conclude that

x(0) = 0

x'(0) = -8 m/s

The final solution, thus

x(t) = -8sin2t

4 0
2 years ago
A large fraction of the ultraviolet (UV) radiation coming from the sun is absorbed by the atmosphere. The main UV absorber in ou
elena55 [62]

Answer:

3.2\times 10^{-7}\ m or 0.32 μm.

Explanation:

Given:

The radiations are UV radiation.

The frequency of the radiations absorbed (f) = 9.38\times 10^{14}\ Hz

The wavelength of the radiations absorbed (λ) = ?

We know that, the speed of ultraviolet radiations is same as speed of light.

So, speed of UV radiation (v) = 3\times 10^8\ m/s

Now, we also know that, the speed of the electromagnetic radiation is related to its frequency and wavelength and is given as:

v=f\lambda

Now, expressing the above equation in terms of wavelength 'λ', we have:

\lambda=\frac{v}{f}

Now, plug in the given values and solve for 'λ'. This gives,

\lambda=\frac{3\times 10^8\ m/s}{9.38\times 10^{14}\ Hz}\\\\\lambda=3.2\times 10^{-7}\ m\\\\\lambda=3.2\times 10^{-7}\times 10^{6}\ \mu m\ [1\ m=10^6\ \mu m]\\\\\lambda=3.2\times 10^{-1}=0.32\ \mu m

Therefore, the wavelength of the radiations absorbed by the ozone is nearly 3.2\times 10^{-7}\ m or 0.32 μm.

7 0
3 years ago
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