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kolbaska11 [484]
3 years ago
14

1. How would the interference pattern change for this experiment if a. the grating was moved twice as far from the screen and b.

the line density of the grating were doubled? Refer to the diffraction grating equations in your answer.
Physics
1 answer:
jolli1 [7]3 years ago
5 0

Answer:

a) The distance between the ineas doubles, intensity decreases with distance

b) The distance between the ineas doubles

Explanation:

The diffraction pattern of a grid is given as a percentage

                d sin θ = m λ

where d give the distance between two consecutive lines, θ it is at an angle, λ the wavelength and m is an integer that determines the order of diffraction, let's not forget that the entire spectrum is at a value of m and then it is repeated.

Let's apply this to our case

a) distance from grid to observation screen doubles

Here we have two effect:

* the energy of the source is constant, it must be distributed over a surface, therefore the intensity decreases with distance

* The other factor can be found using trigonometry

          tan θ= y / L

where y is the distance from the central maximum to the line under study and L is the distance to the screen

           

In general, diffraction experiments cover very small angles

             tan θ = sin θ/ cos θ = sin θ

we substitute

          sinθ = y / L

we subtitle into the diffraction equation

          d y / L = m λ

          y = L / d m λ

          L = 2 L₀

          y = 2 L₀ m λ / d

we see that by doubling the distance to the screen the lines we are seeing are separated by double

b) When the density of lines doubles, it means that in the same distance I have twice as many lines, therefore the distance between two consecutive lines is reduced by half

          d = d₀o / 2

          y = (L m λ) / d

          y = (L m λ/ d₀) 2

we see that The distance between the ineas doubles

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Answer:

C. 2.3A

Explanation:

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3 years ago
The two uniform, slender rods B1and B2, each of mass 2kg, are pinned together at P, and then B1is suspended from a pin at O. (Th
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Answer:

hello the diagram relating to this question is attached below

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b) Force exerted on B2 at P = 39.2 N

Explanation:

Given data:

Co = 150 N-m ,

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at point P ; Co = I* ∝B2'

                150  = ( (2*0.5^2) / 3 ) * ∝B2

∴ ∝B2' = 900 rad/sec

hence angular acceleration of B2 = ∝B2' + ∝B1 = 900 + 180 = 1080 rad/sec

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                  150 = [ (2*0.5^2) / 3  + (2*0.5^2) / 3  + (2*0.5^2) ] * ∝B1

∴ ∝B1 = 180 rad/sec

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T2 = mB1g + T1  -------- ( 1 )

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3 years ago
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A 20-kg block is held at rest on the inclined slope by a peg. A 2-kg pendulum starts at rest in a horizontal position when it is
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Complete Question

The diagram of this question is shown on the first uploaded image

Answer:

The distance the block slides before stopping is d = 0.313 \ m

Explanation:

The free body diagram for the diagram in the question is shown

From the diagram the angle is \theta = 25 ^o

         sin \theta  = \frac{h}{d}

Where h = h_b - h_a

     So      d sin \theta  = h_b - h_a

From the question we are told that

      The mass of the block is  m = 20 \ kg

       The mass of the pendulum is  m_p = 2 \ kg

       The velocity of the pendulum at the bottom of swing is v_p = 15 m/s

        The coefficient of restitution is  e =0.7

         The coefficient of kinetic friction is  \mu _k = 0.5

The velocity of the block after the impact is mathematically represented as

            v_2 f = \frac{m_b - em_p}{m_b + m_p}  * v_2 i + \frac{[1 + e] m_1}{m_1 + m_2 } v_p

Where  v_2 i is the velocity of the block  before collision which is  0

                  = \frac{20 - (0.7 * 2)}{(2 + 20)} * 0 + \frac{(1 + 0.7) * 2 }{2 + 20}   * 15

Substituting value

                   v_2 f = 2.310\  m/s

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      The energy at point a  =  energy at point b

So    PE_A + KE _A = PE_B + KE_B  +  E_F

Where  

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          PE_A = m_b gh_a = 0 at the bottom

      KE _A is the kinetic energy at A  which is mathematically represented as

               K_A = \frac{1}{2} m_b * v_2f^2                  

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            PE_B = m_b gh

From the diagram h = h_b -h_a

       PE_B = m_b g(h_b - h_a)

KE _B is the kinetic energy at B  which is 0 (at the top )

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So

   \frac{1}{2} m_b v_2 f^2  = m_b g h_b + \mu_k m_b g cos \25 * d

Substituting values

   \frac{1}{2}  * 20 * 2.310^2 = 20 * 9.8 * d sin(25)  + 0.5* 20 * 9.8 * cos 25 * d    

So

       d = 0.313 \ m

       

   

6 0
3 years ago
A 3.2-kg thin, circular hoop with a radius of 5.4 m is rotating about an axis through its center and perpendicular to its plane.
elena-14-01-66 [18.8K]

Answer:

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6 0
3 years ago
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