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tia_tia [17]
2 years ago
15

The rhinestones in costume jewelry are glass with index of refraction 1.50. to make them more reflective, they are often coated

with a layer of silicon monoxide of index of refraction 2.00. what is the minimum coating thickness needed to ensure that light of wavelength 560 nm and of perpendicular incidence will be reflected from the two surfaces of the coating with fully constructive interference?
Physics
1 answer:
nevsk [136]2 years ago
7 0
What is he minumum coating of thickness needed to ensure that lifght of waveelntght 5660 mbnd si
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Question 1<br> The orange line is measuring the
11111nata11111 [884]

Answer:

Approximately 4.5 square miles

Explanation:

(surrounding van Nuys and sepulveda stations) and one measuring approximately 2.85 square miles ( surrounding north Hollywood Station)

4 0
3 years ago
Which groups on the periodic table contain metalloids?
N76 [4]
Metalloids are in Group 13 to 16
6 0
3 years ago
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While standing on a bridge 40.0 m above the ground, you drop a stone from rest. when the stone has fallen 3.80 m, you throw a se
Norma-Jean [14]
Given: distance 1 d₁ = 40 m;  distance 2 d₂ = 3.8 m   g = -9.8 m/s²

 Initial Velocity Vi = 0  Final Velocity of stone 2 is unknown = ?

Total distance dₓ = d₁ - d₂  = 40 m - 3.8 m = 36.2 m

Formula: a = Vf² -  Vi²/2d   derive for Final Velocity Vf
 
acceleration is now due to gravity, therefore a = g

Vf = √2gd   Vf = √2(9.8 m/s²)(36.2 m)

Vf = 26.64 m/s  

Reason: The second stone will still start from rest.


3 0
3 years ago
A 59kg child starting from rest slides down a water slide with a vertical height of 5.0m. what is the child's speed halfway down
KIM [24]
<span>EP (potential energy) = mgy -> (59)(9.8)(-5) = -2,891
   EP + EK (kinetic energy) = 0; but rearranging it for EK makes it EK = -EP, such that EK = 2891 when plugged in.
   EK = 0.5mv^2, but can also be v = sqrt(2EK/m).
   Plugging that in for sqrt((2 * 2891)/59), we get 9.9 m/s^2 with respect to significant figures.</span>
6 0
3 years ago
An 67-kg jogger is heading due east at a speed of 2.3 m/s. A 70-kg jogger is heading 61 ° north of east at a speed of 1.3 m/s. F
ololo11 [35]

Answer

given,

mass of jogger  = 67 kg

speed in east direction = 2.3 m/s

mass of jogger 2 = 70 Kg

speed  = 1.3 m/s  in  61 ° north of east.

jogger one

P_1 = m_1 v_1 \hat{i}

P_1 = 67 \times 2.3\hat{i}

P_1 = 154.1 \hat{i}

P_2 = m_2 v_2 \hat{i} +m_2 v_2 \hat{j}

P_2 = 70\times v cos \theta \hat{i} +70\times v sin \theta \hat{j}

P_2 = 70\times 1.3 cos 61^0 \hat{i} +70\times 1.3 sin 61^0\hat{j}

P_2 = 44.12\hat{i} +79.59\hat{j}

now

P = P₁ + P₂

P = 198.22 \hat{i} +79.59 \hat{j}

magnitude

P = \sqrt{198.22^2 + 79.59^2}

P =213.60 kg.m/s

\theta = tan^{-1}\dfrac{79.59}{198.22}

\theta = 21.87

the angle is \theta = 21.87 north of east

7 0
3 years ago
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