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Korolek [52]
8 months ago
13

a nearly monochromatic beam of light with an average wavelength of 514 nm is incident normally on a diffraction grating 1.30 cm

wide with 110 lines/cm. what is the smallest wavelength difference (in nm) this grating can resolve in the second order diffraction with this beam?
Physics
1 answer:
muminat8 months ago
8 0

smallest wavelength difference (in nm) this grating can resolve in the second order diffraction with this beam is 8 degree

The formula for diffraction grating: Obviously, d = \frac {1} { N }, where N is the grating constant, and it is the number of lines per unit length. Also, n is the order of grating, which is a positive integer, representing the repetition of the spectrum so in this case the wavelength difference can be calculated as  514 nm/ 1.30 X 110 =3.59 X2 =approx 8.Wavelengths between 100 nm and 10 m correspond to the optical regime, where grating application is most prevalent. In that situation, the groove density can range from a few tens to a few thousand grooves per millimeter, similar to echelle gratings.

to learn more about wavelength

brainly.com/question/13533093

#SPJ4

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In pairs figure-skating competition, a 65-kg man and his 45-kg female partner stand facing each other both at rest on the ice. I
Lina20 [59]

Answer:

The final velocity of her partner is approximately -1.04 m/s or 1.04 m/s in the opposite direction to her direction of motion

Explanation:

The given parameters are;

The mass of the man, m₁ = 65 kg

The mass of the woman, m₂ = 45 kg

Taking the relative initial velocity of the man and the woman as 0 m/s, we have;

The initial velocity of the man, v₁₁ = 0 m/s

The initial velocity of the man, v₁₂ = 0 m/s

The final velocity of the woman, v₂₂ = 1.5 m/s

The final velocity of the man = v₂₁

Therefore, we have, by the conservation of momentum principle;

The total initial momentum = The total final momentum

Which gives;

m₁ × v₁₁ + m₂ × v₁₂ = m₁ × v₂₁ + m₂ × v₂₂

Substituting the known values;

65 × 0 + 45 × 0 = 65 × v₂₁ + 45 × 1.5

∴ 65 × v₂₁ + 45 × 1.5 = 0

45 × 1.5 = - 65 × v₂₁

v₂₁ = 45 × 1.5/(-65) ≈ -1.04 m/s

The final velocity of the man, her partner = v₂₁ ≈ -1.04 m/s.

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you are piloting a small plane and you want to reach an airport 450 km due south in 3.0 h a wind is blowing from the west 50.0 k
alex41 [277]

Answer:

You should choose airspeed 158.11 km/h at 18.4° west of south

Explanation:

The distance to the air port is 450 km due to south

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→ Distance = 450 km , time = 3 hours

→ The velocity of your plane = 450 ÷ 3 = 150 km/h due to south

A wind is blowing from west 50 km/h

We need to know what heading and airspeed you should choose to

reach your destination

At first we must find the resultant velocity of your plane and the wind

The south and west are perpendicular, then the resultant velocity is

→ v_{R}=\sqrt{(v_{p})^{2}+(v_{w})^{2}}

→ v_{p}=150 km/h ,  v_{w}=50 km/h

→ v_{R}=\sqrt{(150)^{2}+(50)^{2}}=158.11 km/h

To cancel the velocity of the wind, the pilot should maintain the velocity

of the plane at 158.11 km/h

The direction of the velocity is the angle between the resultant velocity

and the vertical (south)

→ The direction of the velocity is tan^{-1}\frac{50}{150}=18.4°

The direction of the velocity is 18.4° west of south

<em>You should choose airspeed 158.11 km/h at 18.4° west of south</em>

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