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Mkey [24]
3 years ago
15

A ray moving in plastic at 62.9 deg enters water, where it bends to 70.9 deg. What is the index of refraction of the plastic?

Physics
1 answer:
DedPeter [7]3 years ago
7 0

Answer:

refractive index of plastic is 1.42

Explanation:

When light ray enters from one medium to other medium then due to transition of light it bends away or towards the normal, this phenomenon is known as refraction of light

So here we know that

n_1 sin i = n_2 sin r

here we have

n_2 = \frac{4}{3}

i = 62.9^o

r = 70.9^o

now we have

n_1 sin62.9 = (\frac{4}{3}) sin70.9

n_1 = 1.42

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Elanso [62]

Answer:

v_{f} = 0.51 \frac{m}{s}

Explanation:

We apply Newton's second law at the crate :

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Data:

m=90kg :  crate mass

F= 282 N

μk =0.351 :coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

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W= m*g

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μk: coefficient of kinetic friction

N : Normal force (N)  

Problem development

We apply the formula (1)

∑Fy = m*ay    , ay=0

N-W = 0

N = W

N = 882 N

We replace the  data in the formula (2)

Ff= μk*N  = 0.351* 882 N

Ff=  309.58 N

We apply the formula (1) in x direction:

∑Fx = m*ax    , ax=0

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a= - 0.306 m/s²

Kinematics of the crate

Because the crate moves with uniformly accelerated movement we apply the following formula :

vf²=v₀²+2*a*d Formula (3)

Where:  

d:displacement in meters (m)  

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s²

Data

v₀ = 0.850 m/s

d = 0.75 m

a= - 0.306 m/s²

We replace the  data in the formula (3)

vf²=(0.850)²+(2)( - 0.306 )(0.75 )

v_{f} = \sqrt{(0.850)^{2} +(2)( - 0.306 )(0.75 )}

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Answer:

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Explanation:

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Musya8 [376]

Answer:

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Explanation:

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erastova [34]

Answer:

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Explanation:

3 0
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