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Pepsi [2]
3 years ago
6

A tsunami, an ocean wave generated by an earthquake, propagates along the open ocean at 700 km/hr and has a wavelength of 750 km

. What is the frequency of the waves in such a tsunami?A. 6.8 HzB. 0.93 HzC. 0.00026 HzD. 1.1 HzE. 0.15 Hz
Physics
1 answer:
Colt1911 [192]3 years ago
7 0

Answer:

C) 0.00026 Hz

Explanation:

  • In any wave, there exists a fixed relationship between the speed v, the frequency f and the wavelength λ, as follows:

       v = \lambda * f (1)

  • Replacing by the givens in (1), and solving for f, we get:

       f = \frac{v}{\lambda} = \frac{700km/hr}{750 km} = 0.93 1/hr (2)

  • Converting the units to Hz (1/sec), we get:

       f = 0.93 \frac{1}{hr} *\frac{1 hr}{3600sec} = 2.6e-4 = 0.00026 Hz (2)

  • The answer C. is the right one.
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What is the power of 10 when 0.00503 is written in scientific notation?
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Negative 3

Explanation:

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Which bone would a forensic anthropologist analyze to identify a victim as male or female?
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A ray of light traveling through air strikes a piece of diamond at an angle of incidence equal to 56 degrees. Calculate the angu
Montano1993 [528]

Answer:

The angle of separation is  \Delta \theta =  0.93 ^o

Explanation:

From the question we are told that

    The angle of incidence is  \theta  _ i  = 56^o

     The refractive index of violet light  in diamond  is  n_v = 2.46

       The refractive index of red light in diamond is n_r = 2.41

      The wavelength of violet light is  \lambda _v = 400nm = 400*10^{-9}m

         The wavelength of red  light is  \lambda _r = 700nm = 700*10^{-9}m

Snell's  Law can be represented mathematically as

         \frac{sin \theta_i}{sin \theta_r} = n

Where \theta_r is the angle of refraction

=>       sin \theta_r  =   \frac{sin \theta_i}{n}

Now considering violet light

               sin \theta_r__{v}}  =   \frac{sin \theta_i}{n_v}

substituting values

                sin \theta_r__{v}}  =   \frac{sin (56)}{2.46}

                 sin \theta_r__{v}}  =  0.337

                 \theta_r__{v}}  =  sin ^{-1} (0.337)

                 \theta_r__{v}}  =  19.69^o

Now considering red light

               sin \theta_r__{R}}  =   \frac{sin \theta_i}{n_r}

substituting values

                sin \theta_r__{R}}  =   \frac{sin (56)}{2.41}

                 sin \theta_r__{R}}  =  0.344

                 \theta_r__{R}}  =  sin ^{-1} (0.344)

                 \theta_r__{R}}  = 20.12^o

The angle of separation between the red light and the violet light is mathematically evaluated as

                  \Delta \theta = \theta_r__{R}} -  \theta_r__{V}}

substituting values

                  \Delta \theta =20.12 - 19.69

                  \Delta \theta =  0.93 ^o

6 0
4 years ago
rickey approaches third base. He dives head first, hitting the ground at 6.75 m/s and reaching the base at 5.91 m/s in 2.5 secon
Gekata [30.6K]

Answer:

15.825 m

Explanation:

t = Time taken = 2.5 s

u = Initial velocity = 6.75 m/s

v = Final velocity = 5.91 m/s

s = Displacement

a = Acceleration

Equation of motion

v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{5.91-6.75}{2.5}\\\Rightarrow a=-0.336\ m/s^2

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{5.91^2-6.75^2}{2\times -0.336}\\\Rightarrow s=15.825\ m

The distance Rickey slides across the ground before touching the base is 15.825 m

4 0
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