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kogti [31]
3 years ago
9

You illuminate the grating in a spectrometer at normal incidence θi=0° with a beam of light that has a wavelength of 6562.8 Å. T

he grating has exactly 600 lines per mm.
a. What is the angular deflection of the first-order diffracted beam?
b. What is the angular deflection of the second-order diffracted beam?
Physics
1 answer:
monitta3 years ago
3 0

Answer:

a) θ₁ = 23.14 ° , b) θ₂ = 51.81 °

Explanation:

An address network is described by the expression

     d sin θ = m λ

Where is the distance between lines, λ is the wavelength and m is the order of the spectrum

The distance between one lines, we can find used a rule of proportions

     d = 1/600

     d = 1.67 10⁻³ mm

    d = 1-67 10⁻³ m

Let's calculate the angle

    sin θ = m λ / d

    θ  = sin⁻¹ (m λ / d)

First order

    θ₁ = sin⁻¹ (1 6.5628 10⁻⁷ / 1.67 10⁻⁶)

    θ₁ = sin⁻¹ (3.93 10⁻¹)

    θ₁ = 23.14 °

Second order

     θ₂ = sin⁻¹ (2 6.5628 10⁻⁷ / 1.67 10⁻⁶)

     θ₂ = sin⁻¹ (0.786)

     θ₂ = 51.81 °

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The metal cases of electrical appliances are connected to an earth wire.Which statement is not correct?A The live wire may becom
labwork [276]

Given that metal cases of electrical appliances are connected to the wire, let's select the statement which is not correct from the list of statements.

The earth wire is used to protect you and help reduce the risk of receiving an electric shock. The earth wire reduces the risk of electric shock by creating a path for a fault or lose current to flow to the Earth.

The Live wire may become loose and touch the metal case. In this case, the earth wire will channel the fault current to the earth thereby reducing the risk of electric shock.

If the metal case becomes live, the earth wire conducts current to the ground. This helps prevent electric shock from the metal case.

The Earth wire has low or no resistance. It is always made of copper.

It provides a low resistance path to the ground.

Therefore, the statement ''the earth wire needs to have high resistance'' is NOT current.

ANSWER:

C. The earth wire needs to have a high reistance.

7 0
1 year ago
Describe at least four of the requirements (facts) that any model of solar system formation must include (according to the nebul
kvasek [131]
The nebular theory describes the formation of the solar system and states that the system began as a gigantic cloud of gas and dust called a nebula which eventually condensed to form the sun, planets and other objects in the solar system. The first fact speaks to the formation of the planets, where gravity pulled larger clumps of material closer to form solid rocky planets closer to the sun and gas giants further out. The second requirement is that a nearby explosion or super nova would have to disturb our nebula to trigger rotation and the eventual formation of the sun. The third requirement/fact is that the planets go around the sun in the same direction. the last fact is that the planets go around the sun within 6 degrees of a common plane. This indicates that the solar system formed from a spinning disk of materials.
4 0
3 years ago
Consider the expression below. Assume m is an integer. 6m(2m + 18) Enter an expression in the box that uses the variable m and m
Anika [276]
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8 0
3 years ago
The High Speed Industrial Drill With Diameter Of 98 Cm Develops 5.85hp At 1900 Rpm. What Torque And Force Is Applied To The Dril
STatiana [176]

Answer:

The torque applied by the drill bit is T = 16.2 Nm and the cutting force of the drill bit is F = 33 N.

Explanation:

Given:-

- The diameter of the drill bit, d = 98 cm

- The power at which drill works, P = 5.85 hp

- The rotational speed of drill, N = 1900 rpm

Find:-

What Torque And Force Is Applied To The Drill Bit?

Solution:-

- The amount of torque (T) generated at the periphery of the cutting edges of the drilling bit when it is driven at a power of (P) horsepower at some rotational speed (N).

- The relation between these quantities is given:

                         T = 5252*P / N

                         T = 5252*5.85 / 1900

                         T = 16.171 Nm

- The force (F) applied at the periphery of the drill bit cutting edge at a distance of radius from the center of drill bit can be determined from the definition of Torque (T) being a cross product of the Force (F) and a moment arm (r):

                          T = F*r

Where,   r = d / 2

                          F = 2T / d

                          F = 2*16.171 / 0.98

                          F = 33 N

Answer: The torque applied by the drill bit is T = 16.2 Nm and the cutting force of the drill bit is F = 33 N.

4 0
3 years ago
Read 2 more answers
A 217 Ω resistor, a 0.875 H inductor, and a 6.75 μF capacitor are connected in series across a voltage source that has voltage a
Nataly [62]

For an AC circuit:

I = V/Z

V = AC source voltage, I = total AC current, Z = total impedance

Note: We will be dealing with impedances which take on complex values where j is the square root of -1. All phasor angles are given in radians.

For a resistor R, inductor L, and capacitor C, their impedances are given by:

Z_{R} = R

R = resistance

Z_{L} = jωL

ω = voltage source angular frequency, L = inductance

Z_{C} = -j/(ωC)

ω = voltage source angular frequency, C = capacitance

Given values:

R = 217Ω, L = 0.875H, C = 6.75×10⁻⁶F, ω = 220rad/s

Plug in and calculate the impedances:

Z_{R} = 217Ω

Z_{L} = j(220)(0.875) = j192.5Ω

Z_{C} = -j/(220×6.75×10⁻⁶) = -j673.4Ω

Add up the impedances to get the total impedance Z, then convert Z to polar form:

Z = Z_{R} + Z_{L} + Z_{C}

Z = 217 + j192.5 - j673.4

Z = (217-j480.9)Ω

Z = (527.6∠-1.147)Ω

Back to I = V/Z

Given values:

V = (30.0∠0+220t)V (assume 0 initial phase, and t = time)

Z = (527.6∠-1.147)Ω (from previous computation)

Plug in and solve for I:

I = (30.0∠0+220t)/(527.6∠-1.147)

I = (0.0569∠1.147+220t)A

To get the voltages of each individual component, we'll just multiply I and each of their impedances:

v_{R} = I×Z_{R}

v_{L} = I×Z_{L}

v_{C} = I×Z_{C}

Given values:

I = (0.0569∠1.147+220t)A

Z_{R} = 217Ω = (217∠0)Ω

Z_{L} = j192.5Ω = (192.5∠π/2)Ω

Z_{C} = -j673.4Ω = (673.4∠-π/2)Ω

Plug in and calculate each component's voltage:

v_{R} = (0.0569∠1.147+220t)(217∠0) = (12.35∠1.147+220t)V

v_{L} = (0.0569∠1.147+220t)(192.5∠π/2) = (10.95∠2.718+220t)V

v_{C} = (0.0569∠1.147+220t)(673.4∠-π/2) = (38.32∠-0.4238+220t)V

Now we have the total and individual voltages as functions of time:

V = (30.0∠0+220t)V

v_{R} = (12.35∠1.147+220t)V

v_{L} = (10.95∠2.718+220t)V

v_{C} = (38.32∠-0.4238+220t)V

Plug in t = 22.0×10⁻³s into these values and take the real component (amplitude multiplied by the cosine of the phase) to determine the real voltage values at this point in time:

V = 30.0cos(0+220(22.0×10⁻³)) = 3.82V

v_{R} = 12.35cos(1.147+220(22.0×10⁻³)) = 11.8V

v_{L} = 10.95cos(2.718+220(22.0×10⁻³)) = 3.19V

v_{C} = 38.32cos(-0.4238+220(22.0×10⁻³)) = -11.2V

4 0
3 years ago
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