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MArishka [77]
3 years ago
13

A narrow beam of light containing orange (610 nm) and blue (470 nm) wavelengths goes from polystyrene to air, striking the surfa

ce at a 28.5° incident angle. What is the angle (in degrees) between the colors when they emerge?
Physics
1 answer:
Eva8 [605]3 years ago
6 0

Answer: The angle between the two colours when they emerge is 0.4°

Explanation:

The refracting angle for different colors knowing its refractive index can be calculated using Snell's law.

Ni × sin α = Nr × sinβ

Where Ni is the refractive index for light in incident medium

α is the angle the incident makes with normal

Nr is the refractive index for light in the refractive medium

β is the angle the refracted makes with the normal

Making β the subject

β = arcSin ( Ni × sinα)/Nr

For Orange color

The value of refractive index in polystyrene medium is 1.490 ,that is Ni= 1.490

α is 30° and Nr is the refractive index of air = 1

β = arcSin(1.490×sin30°)/1

β= 48.15°

For Blue color

The value of refractive index in polystyrene medium is 1.499,that is Ni= 1.499

α is 30° and Nr is the refractive index of air = 1

β = arcSin(1.499×sin30°)/1

β= 48.55°

The angle between the two colours is the difference in the angles of their refracted rays

48.55-48.15=0.4°

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Cart A, with a mass of 0.20 kg, travels on a horizontal air trackat 3.0m/s and hits cart B, which has a mass of 0.40 kg and is i
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Answer:

\large \boxed{\text{C. 2.3 m/s}}

Explanation:

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m_{\text{A} } = \text{0.20 kg};\,v_{\text{Ai}} = \text{3.0 m/s}\\m_{\text{B} } = \text{0.40 kg};\,v_{\text{Bi}} = \text{2.0 m/s}\\

Calculation:

This is a perfectly inelastic collision.  The two carts stick together after the collision and move with a common final velocity.

The conservation of momentum equation is

\begin{array}{rcl}m_{\text{A}}v_{\text{Ai}} +m_{\text{B}} v_{\text{Bi}}&=&(m_{\text{A}}  + m_{\text{B}})v_{\text{f}}\\0.20\times 3.0 + 0.40\times 2.0 & = & (0.20 + 0.40)v_{\text{f}}\\0.60 + 0.80 & = & 0.60v_{\text{f}}\\1.40 & = & 0.60v_{\text{f}}\\v_{\text{f}}&=& \dfrac{1.40}{0.60}\\\\& = & \textbf{2.3 m/s}\\\end{array}\\\text{The centre of mass has a velocity of $\large \boxed{\textbf{2.3 m/s}}$}

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¡¡first convert revolutions per second to
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Answer:

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Now r = length of chain = 1.4 m and ω = 0.595 rev/s = 0.595 × 2π/s = 3.74 rad/s.

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The tension in the chain, T = ma where m = mass of hammer = 8.45 kg and a = centripetal acceleration of hammer = 5.23 m/s². This tension is the centripetal force on the hammer.

So, T = 8.45 kg × 5.23 m/s²

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6 0
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