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denis23 [38]
3 years ago
7

A parallel beam of light in air makes an angle of 43.5 ∘ with the surface of a glass plate having a refractive index of 1.68. Yo

u may want to review (Pages 1080 - 1086) . For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of Reflection and refraction.
a. What is the angle between the reflected part of the beam and the surface of the glass? θθ = nothing ∘
b. What is the angle between the refracted beam and the surface of the glass? θθ = nothing ∘
Physics
1 answer:
aniked [119]3 years ago
3 0

Answer:

a) 46.5º  b) 64.4º

Explanation:

To solve this problem we will use the laws of geometric optics

a) For this part we will use the law of reflection that states that the reflected and incident angle are equal

     θ = 43.5º

This angle measured from the surface is

     θ_r = 90 -43.5

     θ_s = 46.5º

b) In this part the law of refraction must be used

     n₁ sin θ₁ = n₂. Sin θ₂

     sin θ₂ = n₁ / n₂ sin θ₁

The index of air refraction is n₁ = 1

The angle is this equation is measured between the vertical line called normal, if the angles are measured with respect to the surface

     θ_s = 90 - θ

     θ_s = 90- 43.5

     θ_s = 46.5º

     sin θ₂ = 1 / 1.68  sin 46.5

     sin θ₂ = 0.4318

     θ₂ = 25.6º

The angle with respect to the surface is

     θ₂_s = 90 - 25.6

     θ₂_s = 64.4º

measured in the fourth quadrant

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1) A uniform wooden beam, with mass of 120 and length L = 4 m, is supported as illustrated in the figure. If the static friction
Kobotan [32]

Answer:

1(a) 55.0°

1(b) 58.3°

2(a) 10.2 N

2(b) 2.61 m/s²

3(a) 76.7°

3(b) 12.8 m/s

3(c) 3.41 s

3(d) 21.8 m/s

3(e) 18.5 m

4(a) 7.35 m/s²

4(b) 31.3 m/s²

4(c) 12.8 m/s²

Explanation:

1) Draw a free body diagram on the beam.  There are five forces:

Weight force mg pulling down at the center of the beam,

Normal force Na pushing up at point A,

Friction force Na μa pushing left at point A,

Normal force Nb pushing perpendicular to the incline at point B,

Friction force Nb μb pushing up the incline at point B.

There are 3 unknown variables: Na, Nb, and θ.  So we're going to need 3 equations.

Sum of forces in the x direction:

∑F = ma

-Na μa + Nb sin φ − Nb μb cos φ = 0

Nb (sin φ − μb cos φ) = Na μa

Nb / Na = μa / (sin φ − μb cos φ)

Sum of forces in the y direction:

∑F = ma

Na + Nb cos φ + Nb μb sin φ − mg = 0

Na = mg − Nb (cos φ + μb sin φ)

Sum of torques about point B:

∑τ = Iα

-mg (L/2) cos θ + Na L cos θ − Na μa L sin θ = 0

mg (L/2) cos θ = Na L cos θ − Na μa L sin θ

mg cos θ = 2 Na cos θ − 2 Na μa sin θ

mg = 2 Na − 2 Na μa tan θ

Substitute:

Na = 2 Na − 2 Na μa tan θ − Nb (cos φ + μb sin φ)

0 = Na − 2 Na μa tan θ − Nb (cos φ + μb sin φ)

Na (1 − 2 μa tan θ) = Nb (cos φ + μb sin φ)

1 − 2 μa tan θ = (Nb / Na) (cos φ + μb sin φ)

2 μa tan θ = 1 − (Nb / Na) (cos φ + μb sin φ)

Substitute again:

2 μa tan θ = 1 − [μa / (sin φ − μb cos φ)] (cos φ + μb sin φ)

tan θ = 1/(2 μa) − (cos φ + μb sin φ) / (2 sin φ − 2 μb cos φ)

a) If φ = 70°, then θ = 55.0°.

b) If φ = 90°, then θ = 58.3°.

2) Draw a free body diagram of each mass.  For each mass, there are four forces.  For mass A:

Weight force Ma g pulling down,

Normal force Na pushing perpendicular to the incline,

Friction force Na μa pushing parallel down the incline,

Tension force T pulling parallel up the incline.

For mass B:

Weight force Mb g pulling down,

Normal force Nb pushing perpendicular to the incline,

Friction force Nb μb pushing parallel up the incline,

Tension force T pulling up the incline.

There are four unknown variables: Na, Nb, T, and a.  So we'll need four equations.

Sum of forces on A in the perpendicular direction:

∑F = ma

Na − Ma g cos θ = 0

Na = Ma g cos θ

Sum of forces on A up the incline:

∑F = ma

T − Na μa − Ma g sin θ = Ma a

T − Ma g cos θ μa − Ma g sin θ = Ma a

Sum of forces on B in the perpendicular direction:

∑F = ma

Nb − Mb g cos φ = 0

Nb = Mb g cos φ

Sum of forces on B down the incline:

∑F = ma

-T − Nb μb + Mb g sin φ = Mb a

-T − Mb g cos φ μb + Mb g sin φ = Mb a

Add together to eliminate T:

-Ma g cos θ μa − Ma g sin θ − Mb g cos φ μb + Mb g sin φ = Ma a + Mb a

g (-Ma (cos θ μa + sin θ) − Mb (cos φ μb − sin φ)) = (Ma + Mb) a

a = -g (Ma (cos θ μa + sin θ) + Mb (cos φ μb − sin φ)) / (Ma + Mb)

a = 2.61 m/s²

Plug into either equation to find T.

T = 10.2 N

3i) Given:

Δx = 3.7 m

vᵧ = 0 m/s

aₓ = 0 m/s²

aᵧ = -10 m/s²

t = 1.25 s

Find: v₀ₓ, v₀ᵧ

Δx = v₀ₓ t + ½ aₓ t²

3.7 m = v₀ₓ (1.25 s) + ½ (0 m/s²) (1.25 s)²

v₀ₓ = 2.96 m/s

vᵧ = aᵧt + v₀ᵧ

0 m/s = (-10 m/s²) (1.25 s) + v₀ᵧ

v₀ᵧ = 12.5 m/s

a) tan θ = v₀ᵧ / v₀ₓ

θ = 76.7°

b) v₀² = v₀ₓ² + v₀ᵧ²

v₀ = 12.8 m/s

3ii) Given:

Δx = D cos 57°

Δy = -D sin 57°

v₀ₓ = 2.96 m/s

v₀ᵧ = 12.5 m/s

aₓ = 0 m/s²

aᵧ = -10 m/s²

c) Find t

Δx = v₀ₓ t + ½ aₓ t²

D cos 57° = (2.96 m/s) t + ½ (0 m/s²) t²

D cos 57° = 2.96t

Δy = v₀ᵧ t + ½ aᵧ t²

-D sin 57° = (12.5 m/s) t + ½ (-10 m/s²) t²

-D sin 57° = 12.5t − 5t²

Divide:

-tan 57° = (12.5t − 5t²) / 2.96t

-4.558t = 12.5t − 5t²

0 = 17.058t  − 5t²

t = 3.41 s

d) Find v

vₓ = aₓt + v₀ₓ

vₓ = (0 m/s²) (3.41 s) + 2.96 m/s

vₓ = 2.96 m/s

vᵧ = aᵧt + v₀ᵧ

vᵧ = (-10 m/s²) (3.41 s) + 12.5 m/s

vᵧ = -21.6 m/s

v² = vₓ² + vᵧ²

v = 21.8 m/s

e) Find D.

D cos 57° = 2.96t

D = 18.5 m

4) Given:

R = 90 m

d = 140 m

v₀ = 0 m/s

at = 0.7t m/s²

The distance to the opposite side of the curve is:

140 m + (90 m) (π/2) = 281 m

a) Find Δx and v if t = 10.5 s.

at = 0.7t

Integrate:

vt = 0.35t² + v₀

vt = 0.35 (10.5)²

vt = 38.6 m/s

Integrate again:

Δx = 0.1167 t³ + v₀ t + x₀

Δx = 0.1167 (10.5)³

Δx = 135 m

The car has not yet reached the curve, so the acceleration is purely tangential.

at = 0.7 (10.5)

at = 7.35 m/s²

b) Find Δx and v if t = 12.2 s.

at = 0.7t

Integrate:

vt = 0.35t² + v₀

vt = 0.35 (12.2)²

vt = 52.1 m/s

Integrate again:

Δx = 0.1167 t³ + v₀ t + x₀

Δx = 0.1167 (12.2)³

Δx = 212 m

The car is in the curve, so it has both tangential and centripetal accelerations.

at = 0.7 (12.2)

at = 8.54 m/s²

ac = v² / r

ac = (52.1 m/s)² / (90 m)

ac = 30.2 m/s²

a² = at² + ac²

a = 31.3 m/s²

c) Given:

Δx = 187 m

v₀ = 0 m/s

at = 3 m/s²

Find: v

v² = v₀² + 2aΔx

v² = (0 m/s)² + 2 (3 m/s²) (187 m)

v = 33.5 m/s

ac = v² / r

ac = (33.5 m/s)² / 90 m

ac = 12.5 m/s²

a² = at² + ac²

a = 12.8 m/s²

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