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Ede4ka [16]
3 years ago
15

Assuming that the limits of the visible spectrum are approximately 380 and 700 nm, find the angular range of the first-order vis

ible spectrum produced by a plane grating that has 900 grooves/mm with the light incident normally on the grating.
Physics
1 answer:
ch4aika [34]3 years ago
3 0

Answer:

angular range is ( 0.681 rad , 0.35 rad )

Explanation:

given data

wavelength λ = 380 nm = 380 × 10^{-9} m

wavelength λ  = 700 nm =  700 × 10^{-9} m

to find out

angular range of the first-order

solution

we will apply here slit experiment equation that is

d sinθ = m λ    ...........1

here m is 1 for single slit and d is = \frac{1}{900*10^3 m}

so put here value in equation 1 for 380 nm

we get

d sinθ = m λ

\frac{1}{900*10^3} sinθ = 1 × 380 × 10^{-9}

θ = 0.35 rad

and for 700 nm

we get

d sinθ = m λ

\frac{1}{900*10^3} sinθ = 1 × 700 × 10^{-9}

θ = 0.681 rad

so angular range is ( 0.681 rad , 0.35 rad )

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Zarrin [17]

Answer:

This is greater than the initial charge, which violates the principle that the charge cannot be created or destroyed, consequently this distribution is impossible to achieve

Explanation:

The metals distribute the charge on all surface when they touch the surface increases so that charge density decreases and when the charge is separated into smaller in each metal.

Let's apply this principle to our case.

One of the spheres is loaded with a charge q, when touching a ball its charge is reduced to 1 / 2q for each ball.

         qA = ½ q

         qB = ½ q

         qC = 0

The total charge is q

we make a second contact

If we touch the ball A again with the other sphere not charged C, the chare is distributed and when separated it is reduced by half

         qA = 1/2 (q / 2) = ¼ q

         qC = ¼ q

         qB = ½ q

At this point all spheres have a charge,

      qA = ¼ q

      qb = ½ q

      qC = ¼ q

The total charge is q

Now let's contact spheres B and one of the other two

       Q = ½ q + ¼ q = ¾ q

When splitting the charge

        qB = ½ ¾ q = 3/8 q

        qC = ½ ¾ q = 3/8 q

        qA = ¼ q

The total charge is q

Note that the total load is always equal to q

Now let's analyze the given configuration

Let's look for the total load

       Q = qA + QB + QC

       Q = ½ q + 3/8 q + ¼ q

        Q = 9/8 q

This is greater than the initial charge, which violates the principle that the charge cannot be created or destroyed, consequently this distribution is impossible to achieve

8 0
2 years ago
What form of energy is a bonfire and a bunsen burner?
xenn [34]

Answer:

heat and light energy

Explanation:

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Gases are more prone to expansion and contraction than liquid . The biggest change in the volume in your thermometer was probabl
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Answer:

In the air

Explanation:

There are three states of matter:

- Solids: in solids, the particles are tightly bond together by strong intermolecular forces, so they cannot move freely - they can only vibrate around their fixed position

- Liquids: in liquids, particles are more free to move, however there are still some intermolecular forces keeping them close to each other

- Gases: in gases, particles are completely free to move, as the intermolecular forces between them are negligible

For this reason, it is generally easier to compress/expand the volume of a gas with respect to the volume of a liquid.

In this problem, we are comparing water (which is a liquid) with air (which is a gas). From what we said above, this means that the change in volume is larger in the air rather than in the water.

8 0
2 years ago
A roller coaster track is 3000 meters long. It takes 100 seconds to travel once around the roller coaster. What is the average s
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All you have to do is divide 3000 by 100, Its 30
6 0
3 years ago
Read 2 more answers
A 25.0-kg child is standing at the edge of a horizontal merry-go-round with a radius of 2.40 m and a moment of inertia of 356 kg
motikmotik

Answer:

\omega_{f}=1.634\ rad/s  

Explanation:

given,  

diameter of merry - go - round = 2.40 m  

moment of inertia = I = 356 kg∙m²

speed of the merry- go-round = 1.80 rad/s

mass of child = 25 kg  

initial angular momentum of the system  

L_i = I\omega_i  

L_i =356\times 1.80  

L_i =640.8\ kg.m^2/s  

final angular momentum of the system  

L_f = (I_{disk}+mR^2)\omega_{f}  

L_f = (356 + 25\times 1.2^2)\omega_{f}  

L_f= (392)\omega_{f}  

from conservation of angular momentum  

L_i = L_f  

640.8= (392)\omega_{f}  

\omega_{f}=1.634\ rad/s  

8 0
2 years ago
Read 2 more answers
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