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Luden [163]
2 years ago
11

In example 18. 4 of the text, the deflection angle of the laser beam as it exits the prism is 22. 6º. If the prism had been made

of glass instead of polystyrene plastic, what would the deflection angle have been?.
Physics
1 answer:
ankoles [38]2 years ago
6 0

The deflection angle would have been 37.29º if the prism had been made of polystyrene plastic.

As we know that the deflection angle is an angle between the moving object from its directed course.

In the following question the deflection angle as laser enters the beam is, B = 22.6º

refractive index of glass, n1= 1.52

refractive index of polystyrene plastic, n2 = 1.59

Now when we see for the second surface then ,

B  = 45º - 22.6º

B  = 22.40º

So from Snell's law:

n1 sinθ1 = n2 sinθ2

→ sin B / sinθ = n1 / n2

sinθ = (sin 22.40º) × ( 1/ 1.59)

θ = 37.29º

Thus, the deflection angle should be 37.29º if the prism is made of polystyrene plastic.

Learn more about deflection angle here:

brainly.com/question/22953155

#SPJ4

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Let's use Newton's Law of Second Motion: F=ma. When no other direct force is acting on the system, the acceleration is due to the gravity. The modified equation becomes: F = mg. So, yes, you need to take into account the gravitational accelerations in the moon and on Earth. 

g,moon = 1.622 m/s²
g,Earth = 9.81 m/s²

The net force is the tension of the string:

F,Earth - F,moon = Tension
Tension = (1/1000 kg)(9.81 m/s²) - (1/1000 kg)(1.622 m/s²)
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To convert, 1 pound force is equal to 4.45 Newtons:

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3 years ago
What is the name of the system of measurement used by most countries around the world?
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Add a vector whose magnitude is 13 with angle 27 degrees to one whose magnitude is 11 with angle 45 degrees? Put the length firs
mafiozo [28]

Answer:

Magnitude of the vector is 23.75\ \text{units} and the direction is 35.23^{\circ}

Explanation:

Magnitude of first vector = |A| = 13\ \text{units}

Angle = \theta_1=27^{\circ}

Magnitude of second vector = |B| = 11\ \text{units}

Angle = \theta_2=45^{\circ}

x component of first vector

A_{x}=|A|\cos\theta_1\\\Rightarrow A_x=13\cos27^{\circ}\\\Rightarrow A_x=11.6\ \text{units}

y component of first vector

A_{y}=|A|\sin\theta_1\\\Rightarrow A_y=13\sin27^{\circ}\\\Rightarrow A_y=5.9\ \text{units}

x component of second vector

B_{x}=|B|\cos\theta_2\\\Rightarrow B_x=11\cos45^{\circ}\\\Rightarrow B_x=7.8\ \text{units}

y component of first vector

B_{y}=|B|\sin\theta_2\\\Rightarrow B_y=11\sin45^{\circ}\\\Rightarrow A_y=7.8\ \text{units}

Adding the magnitudes

C_x=A_x+B_x=11.6+7.8\\\Rightarrow C_x=19.4\ \text{units}

C_y=A_y+B_y=5.9+7.8\\\Rightarrow C_y=13.7\ \text{units}

Magnitude of the sum of the vectors would be

|C|=\sqrt{C_x^2+C_y^2}\\\Rightarrow |C|=\sqrt{19.4^2+13.7^2}=23.75\ \text{units}

The direction would be

\theta=\tan^{-1}\dfrac{C_y}{C_x}\\\Rightarrow \theta=\tan^{-1}\dfrac{13.7}{19.4}\\\Rightarrow \theta=35.23^{\circ}

The magnitude of the vector is 23.75\ \text{units} and the direction is 35.23^{\circ}

4 0
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One day, after pulling down your window shade, you notice that sunlight is passing through a pinhole in the shade and making a s
baherus [9]

Given that,

Central maximum = 1 cm

Distance from the window shade to the wall =4 m

We know that,

The visible range of the sun light is 400 nm to 700 nm.

(a). We need to calculate the average wavelength

Using formula of average wavelength

\lambda_{avg}=\dfrac{\lambda_{1}+\lambda_{2}}{2}

Put the value into the formula

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(b). We need to calculate the diameter of the pinhole

Using formula for diameter

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D=\dfrac{2.44\lambda L}{w}

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D=\dfrac{2.44\times550\times10^{-9}\times4}{1\times10^{-2}}

D=0.537\ mm

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(b). The diameter of the pinhole is 0.537 mm.

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