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LenaWriter [7]
3 years ago
7

A piece of curved glass has a radius of curvature of r = 10.8 m and is used to form Newton's rings, as in the drawing. Not count

ing the dark spot at the center of the pattern, there are one hundred dark fringes, the last one being at the outer edge of the curved piece of glass. The light being used has a wavelength of 652 nm in vacuum. What is the radius R of the outermost dark ring in the pattern? (Hint: Note that r is much greater than R, and you may assume that tan(θ) = θ for small angles, where θ must be expressed in radians.)
Physics
2 answers:
Sergeeva-Olga [200]3 years ago
8 0

Answer:

The Radius is 2.65cm

Explanation:

MAVERICK [17]3 years ago
7 0

Answer:

Radius of the outer most dark fringe is 2.65 cm

Solution:

As per the question:

Radius of curvature of the glass, r = 10.8 m

No. of dark fringes, n = 100

Wavelength of light, \lambda = 652\ nm = 652\times 10^{- 9}\ m

Now,

To calculate the radius R of the outermost ring:

Radius of the dark fringe of nth order is given by:

R^{2} = nr\lambda = 100\times 10.8\times 652\times 10^{- 9} = 7.042\times 10^{- 4}

R = \sqrt{7.042\times 10^{- 4}} = 0.0265\ m = 2.65\ cm

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A certain fm radio station broadcasts electromagnetic waves at a frequency of 125000 hertz. these radio waves travel at a speed
faltersainse [42]
Speed = frequency x wavelength
300,000,000 = 125000 x wavelength
wavelength = 125000/300,000,000 =4.16667x10^-4 meters
or 4.1667E-4 meters
4 0
3 years ago
In a choir practice room, two parallel walls are 5.70 m apart. The singers stand against the north wall. The organist faces the
ololo11 [35]

Answer:

4.98 m

Explanation:

Given that

Width of the mirror, d = 0.6 m

Organist distance to the mirror, s = 0.78 m

Distance between the singer and the organist, S = 5.7 + 0.78 = 6.48 m

Width of north wall, D?

Using the simple relationship

D/S = d/s, on rearranging

D = dS /s

D = (0.6 * 6.48) / 0.78

D = 3.888 / 0.78

D = 4.98 m

Therefore, we can conclude that the Width of north wall is 4.98 m

7 0
2 years ago
2. What makes Newton's 3rd Law an unbalanced force?​
LiRa [457]

Answer:

Explanation:

a

5 0
2 years ago
A person pushes a refrigerator across a horizontal floor. The mass of the refrigerator is 110 kg, the coefficient of static fric
Law Incorporation [45]

Answer:

Explanation:

mass of refrigerator, m = 110 kg

coefficient of static friction, μs = 0.85

coefficient of kinetic friction, μk = 0.59

(a) the minimum force required to just start the motion in refrigerator

F = μs x mg

F = 0.85 x 110 x 9.8

F = 916.3 N

(b) The force required to move the refrigerator with constant speed

F' = μk x mg

F' = 0.59 x 110 x 9.8

F' = 636.02 N

(c) Let a be the acceleration.

Net force = Applied force - friction force

F net = 950 - 636.02

F net = 313.98 N

a = F net / mass

a = 313.98 / 110

a = 2.85 m/s²

4 0
3 years ago
An archer shoots an arrow at a 75.0 m distant target; the bull’s-eye of the target is at same height as the release height of th
jeyben [28]

Answer:

the shooting angle ia 18.4º

Explanation:

For resolution of this exercise we use projectile launch expressions, let's see the scope

      R = Vo² sin (2θ) / g

      sin 2θ = g R / Vo²

      sin 2θ = 9.8 75/35²

      2θ = sin⁻¹ (0.6)

      θ = 18.4º

To know how for the arrow the tree branch we calculate the height of the arrow at this point

       X2 = 75/2 = 37.5 m

We calculate the time to reach this point since the speed is constant on the X axis

       X = Vox t

       t2 = X2 / Vox = X2 / (Vo cosθ)

        t2 = 37.5 / (35 cos 18.4)

        t2 = 1.13 s

With this time we calculate the height at this point

        Y = Voy t - ½ g t²

        Y = 35 sin 18.4   1.13 - ½ 9.8 1,13²

        Y = 6.23 m

With the height of the branch is 3.5 m and the arrow passes to 6.23, it passes over the branch

8 0
3 years ago
Read 2 more answers
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