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Svet_ta [14]
3 years ago
8

A beam of light is emitted 8.6 cm beneath the surface of a liquid and strikes the air surface 7.2 cm from the point directly abo

ve the source. Part A If total internal reflection occurs, what can you say about the minimum possible index of refraction of the liquid?
Physics
1 answer:
Makovka662 [10]3 years ago
4 0

Answer:

n = 1.56

Explanation:

The total reflection attempts occurs when a light beam passes from a medium with a higher index to a medium with a lower nest, at an angle where it occurs we can find them by the refractive relationship

             n₁ sin θ₁ = n₂ sin θ₂

             n1 = n2 / sin θ₁

For this relationship to be fulfilled, the liquid index must be greater than the air index divided by the sine of the critical angle

Let's use trigonometry to find angle

               tan θ = y / x

                θ = tan⁻¹ 7.2 / 8.6

                θ = 39.94º

         

                n₁ = 1 / sin 39.94

                n = 1.56

This is the refractive index of the liquid

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You have found that the net force exerted on the cylinder depends on several different independent variables in different ways.
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The net force on the cylinder depends on several independent variables such as mass, acceleration, velocity, time, etc, and it is given as F(net) = ma.

<h3>Net force exerted on the cylinder</h3>

The net force on the cylinder is calculated from Newton's second law of motion.

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A box of mass m slides down an inclined plane that makes an angle of φ with the horizontal. if the coefficient of kinetic fricti
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4 years ago
A small object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t). At t = 0 the object is at x = -14.0 m an
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Complete Question

A small object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t)−(0.0320m/s3)(15.0s−t). At t = 0 the object is at x = -14.0 m and has velocity v0x = 7.10 m/s. What is the x-coordinate of the object when t = 10.0 s?

Answer:

The position of the object at t = 10s is  X  =  38.3 \  m

Explanation:

From the question we are told that

The acceleration along the x axis is  a_{x}t  =  -(0.0320\ m/s^3)(15.0 s- t)- (0.0320\ m/s^3)

  The position of the object at t = 0 is  x = -14.0 m

  The velocity at t = 0 s is  v_{0}x = 7.10 m/s

Generally from the equation for acceleration along x axis we have that

     a_x = \frac{dV_{x}}{dt}  = -0.032 (15- t)

=>   \int\limits  {dV_{x}} \, = \int\limits  {-0.032(15- t)} \, dt

=>   V_{x} = -0.032 [15t - \frac{t^2 }{2} ]+ K_1

At  t =0  s   and  v_{0}x = 7.10 m/s

=>   7.10  = -0.032 [15(0) - \frac{(0)^2 }{2} ]+ K_1

=>   K_1 = 7.10      

So  

      \frac{dX}{dt}  = -0.032 [15t - \frac{t^2 }{2} ]+ K_1

=>  \int\limits dX  = \int\limits [-0.032 [15t - \frac{t^2 }{2} ]+ K_1] }{dt}

=>  X  =  -0.032 [ 15\frac{t^2}{2}  - \frac{t^3 }{6} ]+ K_1t +K_2

At  t =0  s   and   x = -14.0 m

  -14  =  -0.032 [ 15\frac{0^2}{2}  - \frac{0^3 }{6} ]+ K_1(0) +K_2

=>   K_2 = -14

So

     X  =  -0.032 [ 15\frac{t^2}{2}  - \frac{t^3 }{6} ]+ 7.10 t -14

At  t = 10.0 s

      X  =  -0.032 [ 15\frac{10^2}{2}  - \frac{10^3 }{6} ]+ 7.10 (10) -14

=>   X  =  38.3 \  m

             

     

5 0
3 years ago
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