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Alexandra [31]
3 years ago
14

A diffraction grating is placed 1.00 m from a viewing screen. Light from a hydrogen lamp goes through the grating. A hydrogen sp

ectral line with a wavelength of 656 nm is seen 60.0 cm to one side of the center. Then, the hydrogen lamp is replaced with an unknown lamp. A spectral line is seen on the screen 36.4 cm away from the center. What is the wavelength of this spectral line
Physics
1 answer:
Colt1911 [192]3 years ago
8 0

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

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Why does it takes the outer planets so long to orbit the Sun? Try to come up with two reasons.
tensa zangetsu [6.8K]

well they are normally bigger than the inner planets, and they also have a bigger distance to go


7 0
3 years ago
A cellist tunes the C string of her instrument to a fundamental frequency of 65.4 Hz. The vibrating portion of the string is 0.5
Dahasolnce [82]

Answer:

a

\lambda  = 1.18 \  m

b

v  =  77.172 \  m/s

c

T  = 151.41 \  N

Explanation:

From the question we are told that

   The frequency is  f =  65.4 \  Hz

   The  length of the vibrating string is  L  =  0.590 \  m

   The  mass is  m  =  15.0 \ g  =  0.015 \  kg

Generally the wavelength is mathematically represented as

           \lambda =  2 *  L

=>        \lambda  =  2 *   0.590

=>         \lambda  = 1.18 \  m

Generally the wave speed is  

          v  =  \lambda  *  f

=>       v  =  1.18 * 65.4

=>       v  =  77.172 \  m/s

Generally the tension on the wire is mathematically represented as

        T  =  v^2  *  \frac{ m }{L }

=>      T  =  77.172 ^2  *  \frac{  0.015  }{0.590}

=>      T  = 151.41 \  N

7 0
3 years ago
A −4.00 μC charge sits in static equilibrium in the center of a conducting spherical shell that has an inner radius 3.13 cm and
Mariulka [41]

Answer:

(a). The charge on the outer surface is −2.43 μC.

(b). The charge on the inner surface is 4.00 μC.

(c). The electric field outside the shell is 3.39\times10^{7}\ N/C

Explanation:

Given that,

Charge q₁ = -4.00 μC

Inner radius = 3.13 m

Outer radius = 4.13 cm

Net charge q₂ = -6.43 μC

We need to calculate the charge on the outer surface

Using formula of charge

q_{out}=q_{2}-q_{1}

q_{out}=-6.43-(-4.00)

q_{out}=-2.43\ \mu C

The charge on the inner surface is q.

q+(-2.43)=-6.43

q=-6.43+2.43= 4.00\ \mu C

We need to calculate the electric field outside the shell

Using formula of electric field

E=\dfrac{kq}{r^2}

Put the value into the formula

E=\dfrac{9\times10^{9}\times6.43\times10^{-6}}{(4.13\times10^{-2})^2}

E=33927618.73\ N/C

E=3.39\times10^{7}\ N/C

Hence, (a). The charge on the outer surface is −2.43 μC.

(b). The charge on the inner surface is 4.00 μC.

(c). The electric field outside the shell is 3.39\times10^{7}\ N/C

5 0
3 years ago
ANSWER ASAPPPPPPPP BE 100% CORRECT
AfilCa [17]
Mass of a sample of gas doesn't change, no matter what happens to its pressure, volume, or temperature.
7 0
3 years ago
Happy Valentines Day<br><br> FR33 points for you guys
Fed [463]

Answer:

thankkkkkksssssssssssssss

6 0
3 years ago
Read 2 more answers
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