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JulsSmile [24]
3 years ago
8

A diffraction grating has 2000 lines per centimeter. at what angle will the first-order maximum be for 520-nm-wavelength green l

ight?a diffraction grating has 2000 lines per centimeter. at what angle will the first-order maximum be for 520-nm-wavelength green light?
Physics
1 answer:
oksano4ka [1.4K]3 years ago
8 0
Applying diffraction equations;
d = 0.01/Number of lines; where 0.01 m = 1 cm, and d = spacing between lines

Therefore,
d = 0.01/2000 = 5*10^-6 m

Additionally,
d*Sin x = m*y; where x = Angle, m = order = 1, y = wavelength = 520 nm =520*10^-9 m

Substituting;
Sin x = my/d = (1*520*10^-9)/(5*10^-6) = 0.1040
x = Sin ^-1(0.104) = 5.97°

Therefore, first-order maximum for 520 nm will be 5.97°.
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Mercury has one of the lowest specific heats. This fact added to its liquid state at most atmospheric temperatures make it effec
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The specific heat of mercury is 149.4 J/(kgK)

Explanation:

When a substance is supplied with an amount of energy Q, its temperature increases according to the equation:

\Delta T=\frac{Q}{mC_s}

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\Delta T is the increase in temperature

m is the mass of the sample

C_s is its specific heat capacity

For the sample of mercury in this problem we have

Q = 275 J

m = 0.450 kg

\Delta T = 4.09 K

Therefore, by re-arranging the equation we find the mercury's specific heat:

C_s = \frac{Q}{m\Delta T}=\frac{275}{(0.450)(4.09)}=149.4 J/(kgK)

Learn more about specific heat capacity:

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A large sheet of charge has a uniform charge density of 9  μCm2. What is the electric field due to this charge at a point just
Alex73 [517]

Answer:

Explanation:

Surface charge density, σ = 9 μC/m² = 9 x 10^-6 C/m²

According to the Gauss theorem,

Electric field due to the sheet is given by

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A stone with a mass of 0.100kg rests on a frictionless, horizontal surface. A bullet of mass 2.50g traveling horizontally at 500
jolli1 [7]

Answer:

Explanation:

Given that:

mass of stone (M) = 0.100 kg

mass of bullet (m) = 2.50 g = 2.5  ×10 ⁻³ kg

initial velocity of stone (u_{stone}) = 0 m/s

Initial velocity of bullet (u_{bullet}) = (500 m/s)i

Speed of the bullet after collision (v_{bullet}) = (300 m/s) j

Suppose we represent (v_{stone}) to be the velocity of the stone after the truck, then:

From linear momentum, the law of conservation can be applied which is expressed as:

m*u_{bullet} + M*{u_{stone}}= mv_{bullet}+Mv_{stone}

(2.50*10^{-3} \ kg) (500)i+0 = (2.50*10^{-3} \ kg)(300 \ m/s)j + (0.100 \ kg)v_{stone}

(2.50*10^{-3} \ kg) (500)i- (2.50*10^{-3} \ kg)(300 \ m/s)j=  (0.100 \ kg)v_{stone}

v_{stone}= (1.25\  kg.m/s)i-(0.75\ kg m/s)j

v_{stone}= (12.5\  m/s)i-(7.5\ m/s)j

∴

The magnitude now is:

v_{stone}=\sqrt{ (12.5\  m/s)^2-(7.5\ m/s)^2}

\mathbf{v_{stone}= 14.6 \ m/s}

Using the tangent of an angle to determine the direction of the velocity after the struck;

Let θ represent the direction:

\theta = tan^{-1} (\dfrac{-7.5}{12.5})

\mathbf{\theta = 30.96^0 \ below \ the \ horizontal\ level}

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The North Pole would be your answer
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