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Artemon [7]
4 years ago
9

Why the center of Newton rings is dark?​

Physics
1 answer:
statuscvo [17]4 years ago
5 0

Answer:

Because at the point of contact the path difference is zero but one of the interfering ray is reflected so the effective path difference becomes λ/2 thus the condition of minimum intensity is created hence centre of ring pattern is

Explanation:

Off google not mine

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What happens when an immovable object meets an unstoppable force?”
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Tarzan, who weighs 825 N, swings from a cliff at the end of a 19.7 m vine that hangs from a high tree limb and initially makes a
kodGreya [7K]

Answer:

a) T = (281.47 i ^ + 714.56 j ^) N , b) F_net = (281.47 i ^ - 110.44 j ^) N ,

c)  F = 281.70 N, d)    θ = 338.58º , e)  a = 3,588 m / s² , f)  θ = 201.45º

Explanation:

For this exercise we will use Newton's second law on each axis

X axis

         -Tₓ = m aₓ

Y Axisy

          T_{y} –W = m a_{y}

Let's use trigonometry to find the components of force

          sin 21.5 = Tₓ / T

          cos 21.5 = T_{y} / T

          Tₓ = T sin 21.5

          T_{y} = T cos 21.5

          Tₓ = 768 sin 21.5 = 281.47 N

          T_{y} = 768 cos 21.5 = 714.56 N

a) the force of the rope on Tarzan is

          T = (281.47 i ^ + 714.56 j ^) N

b) The net force is the subtraction of the tension minus the weight of Tarzan

Y  Axis   F_net = 714.56 - 825 = -110.44 N

              F_net = (281.47 i ^ - 110.44 j ^) N

c) Let's use Pythagoras' theorem

      F = √ (Fₓ² + T_{y}²)

      F = √ (281.47² + 110.44²)

      F = 281.70 N

d) Let's use trigonometry

     tan θ = F_{y} / Fₓ

      θ = tan⁻¹ F_{y} / Fₓ

      θ = tan⁻¹ (-110.44 / 281.47)

       θ = -21.42º

This angle is average clockwise, for counterclockwise measurement

       θ = 360 - 21.42

       θ = 338.58º

Acceleration

X axis

             Tₓ = m aₓ

             aₓ = Tₓ / m

The mass of Tarzan is

             m = W / g

             m = 825 / 9.8 = 84.18 kg

             

             aₓ = 281.47 / 84.18

             aₓ = -3.34 m / s2

Y Axis

            T_{y}-W = m a_{y}

            a_{y} = (T_{y} -W) / m

            a_{y} = (714.56-825) / 84.18

            a_{y} = - 1,312 m / s²

Acceleration Module

             a = √ aₓ² + a_{y}²

             a = √ (3.34² +1.312²)

             a = 3,588 m / s²

The angle

          θ = tan⁻¹ a_{y} / aₓ

          θ = tan⁻¹ (-1312 / -3.34)

          θ = 21.45º

Notice that the two components of the acceleration are negative, so the angle is in the third quadrant, to measure from the x-axis

          θ = 180 + 21.45

          θ = 201.45º

3 0
3 years ago
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