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Karo-lina-s [1.5K]
3 years ago
11

spotlight on a boat is 2.5 m above the water, and the light strikes the water at a point that is 8.0 m horizontally displaced fr

om the spotlight (see the drawing). The depth of the water is 4.0 m. Determine the distance d, which locates the point where the light strikes the bottom. Assume water has a refraction index of 1.333
Physics
1 answer:
ludmilkaskok [199]3 years ago
3 0

Answer:

Explanation:

Let i be the angle of incidence and r be the angle of refraction .

From the figure

Tan ( 90 - i ) = 2.5 / 8

cot i = 2.5 / 8

Tan i = 8 / 2.5 = 3.2

i = 72.65°

From snell's law

sini / sin r = refractive index

sin 72.65 / sinr = 1.333

sin r = .9545  / 1.333

= .72

r = 46⁰

From the figure

Tan r  = d / 4

Tan 46 = d /4

d = 4 x Tan 46

= 4 x 1.0355

=4.14 m .

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mass of the first object, m₁ = 3.35 kg

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mass of the second object, m₂ = 1.88 kg

initial velocity of the second object, u₂ = 3.12 m/s in negative y-direction

initial momentum of the first object, P₁ = 3.35 x 4.9 = 16.415 kgm/s

initial momentum of the second object, P₂ = 1.88 x 3.12 = 5.8656 kgm/s

The resultant velocity of the two objects is given by;

R² = 16.415² + 5.8656²

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R = √303.858

R = 17.432 kgm/s

Apply the principle of conservation of linear momentum for inelastic collision;

total initial momentum before = total final momentum after collision

P₁(x) + P₂(y) = Pf

R = Pf

R = v(m₁ + m₂)

17.432 = v(m₁ + m₂)

where;

v is the final components of velocity of the composite object

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Therefore, the final components of velocity of the composite object is 3.33 m/s.

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