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amm1812
3 years ago
8

The Sun appears at an angle of 48.4° above the horizontal as viewed by a dolphin swimming underwater. What angle does the sunlig

ht striking the water actually make with the horizon? (Assume nwater = 1.333. Enter an answer between 0° and 90°.)
Physics
1 answer:
Luden [163]3 years ago
6 0

Answer:

4.57 degree

Explanation:

As the dolphin is under water, so the angle is angle of refraction which is equal to 48.4 degree.

refractive index of water, n = 1.333

Let the angle of incidence which is the angle between the incident ray and the normal is i.

By use of Snell's law

n = \frac{Sin i}{Sin r}

1.333 = \frac{Sin i}{Sin 48.4}

Sin i = 1.333\times Sin 48.4 = 0.9968

i = 85.43 degree

Angle made with the horizon = 90 - 85.43 = 4.57 degree

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(10 points) Consider a single crystal of some hypothetical metal that has the FCC crystal structure, and its critical resolved s
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Complete Question

(10 points) Consider a single crystal of some hypothetical metal that has the FCC crystal structure, and its critical resolved shear stress is 3.42 MPa. The crystal is oriented such that a tensile stress is applied at an angle of 39.2 degrees to the slip plane. Slip can occur in two directions( 18.4 and 74.2 degrees to the tensile force).

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b)  (5 points)  What is the yield strength of this crystal during this tensile test ?

Answer:

a

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b

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Explanation:

From the question we are told that

    The Resolved shear force is \sigma = 3.2MPa

     The angle in which the tensile stress is applied is \O = 39.2^o

      The first direction of slip is \theta _ 1 =18.4^o

      The second direction of the slip is \theta_2 = 74.2^o

Generally the condition for direction in which slip is likely to occur(

direction in which slip is favored )   is that  (cos( \O) cos (\theta) )

Must be Maximum for that direction

   Since Cos (\O) is constant for both direction we would  look at the the cos of the the angle for both direction

      Cos (\theta_ 1) = Cos(18.4^o) =0.9488

      Cos (\theta_ 2) = Cos(74.2^o) =0.2722

From this calculation we can see that the slip would occur 18.4° to the tensile force

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           \sigma = \sigma_y * (cos(\O) cos(\theta_1))

Where \sigma_y is the  yield strength

Making  \sigma_y the subject

       \sigma_y = \frac{\sigma }{[cos (\O) cos(\theta_1)]}

Substituting value

             \sigma_y = \frac{3.4*10^{6}}{cos (39.2) (cos 18.4)}

                 \sigma_y = 4.65\  MPa

         

   

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Answer:

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