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Veronika [31]
3 years ago
12

In an experiment conducted using the Young's double-slit apparatus, the separation between the slits is 20 µm. A first-order con

structive interference fringe appears at an angle of 2.5o from the zeroth order (central) fringe.
A. What wavelength of light is used in the experiment?
B. At what angle would the fourth-order (m = 4) bright fringe appear?
C. At what angle would the fourth-order (m=4) dark fringe appear?
Physics
1 answer:
jeka57 [31]3 years ago
6 0
To solve this problem, we use the formula
λ = s sin θ
where s is the separation and θ is the angle interference
So,
λ = 20 x 10^-6 sin 2.5
λ = 8.72 x 10^-7 m

The required angle for the fourth order bright fringe is
θb = sin⁻¹ (4λ / s) = sin⁻¹ (4 (8.72 x 10^-7 m)/ 20 x 10^-6 ) = 10.04°

The required angle for the fourth order dark fringe is
θd = sin⁻¹ (4.5 λ / s) = sin⁻¹ (4.5 (8.72 x 10^-7 m)/ 20 x 10^-6 ) = 11.31°
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Answer:

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In this exercise the change in moment of a ball is asked in two different cases

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b) in this case we have a bouncing ball, this is an elastic collision, as the gate is fixed it can be considered an object of infinite mass, therefore the final speed of the ball has the same modulus of the initial velocity, but address would count

         v_f = - v₀

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        Δp = m v_f - m v₀

        Δp = m (v_f -v₀)

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A bullet is fired with a velocity of 100 m/s from the ground at an angle of 60° with the horizontal. Calculate the horizontal ra
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1) The horizontal range of the bullet is 884 m

2) The maximum height attained by the bullet is 383 m

Explanation:

1)

The motion of the bullet is a projectile motion, which consists of two separate motions:  

- A uniform motion (constant velocity) along the horizontal direction  

- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction  

From the equation of motions along the x- and y- directions, it is possible to find an expression for the horizontal range covered by a projectile, and it is:

d=\frac{u^2 sin 2\theta}{g}

where

u is the initial speed of the projectile

\theta is the angle of projection

g=9.8 m/s^2 is the acceleration of gravity

For the bullet in the problem, we have

u = 100 m/s (initial speed)

\theta=60^{\circ} (angle)

Solving the equation, we find the horizontal range:

d=\frac{(100)^2sin(2\cdot 60^{\circ})}{9.8}=884 m

2)

To find the maximum height, we have to analyze the vertical motion of the bullet. We can do it by using the following suvat equation:

v_y^2 - u_y^2 = 2as

where

v_y is the vertical velocity of the bullet after having covered a vertical displacement of s

u_y is the initial vertical velocity

a =-g= is the acceleration (negative, since it points downward)

The vertical component of the initial velocity is given by

u_y = u sin\theta

Also, the maximum height s is reached when the vertical velocity becomes zero,

v_y =0

Substituting into the equation and re-arranging for s, we find the maximum height:

s=\frac{u^2 sin^2 \theta}{2g}=\frac{(100)^2(sin 60^{\circ})^2}{2(9.8)}=383 m

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

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