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Bess [88]
3 years ago
6

A beam of light strikes a mirror reflects off at the angle in which it hit the mirror. which term does the diagram below illustr

ate A. Rotation B. Reflection C.refraction D.revolution
Physics
1 answer:
zavuch27 [327]3 years ago
3 0

Answer:

B.) reflection

Explanation:

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A water wave has a speed of 22.0 meters/second if the wave frequency is 0.0680 hertz what is the wavelength
serg [7]
V = f(wavelength)
22.0 = 0.0680 (wavelength)
wavelength = 323.52 m
8 0
3 years ago
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
The y-position of a damped oscillator as a function of time is shown in the figure.
NISA [10]

(1) The period of the oscillator is 1 second.

(2) The damping coefficient is 0.93.

<h3>What is period of oscillation?</h3>

The period of oscillation is the time taken to make one complete cycle.

From the graph, the time taken to make one complete oscillation is 1 second.

<h3>Damping coefficient</h3>

equation of the wave is given as;

y(t) = Ae^(-btx) cos(ωt)

<h3>at time, t = 0, y = 3.5</h3>

3.5 = Ae^(-0) cos(0)

3.5 = A x 1

A = 3.5 cm

<h3>at time, t = 1 cm, y = - 3cm</h3>

-3 = 3.5e^(-bx) cos(ω)

-3/3.5 = e^(-bx) cos(ω)

-0.857 = e^(-bx) cos(ω)

-0.857 / cos(ω) =  e^(-bx)

ln[-0.857 / cos(ω)] = -bx  

ln[-0.857 / cos(ω)] / b = - x  ---- (1)

<h3>at time, t = 2 cm, y = - 2cm</h3>

-2 = 3.5e^(-2bx) cos(2ω)

-0.57 = e^(-2bx) cos(2ω)

ln[-0.57 / cos(2ω)] = -2bx  

ln[-0.57 / cos(2ω)] /2b = - x  ------(2)

solve (1) and (2)

ln[-0.57 / cos(2ω)]/2b = ln[-0.857 / cos(ω)] /b

-0.57 / cos(ω) = 2(-0.857 / cos(ω))

2(-0.857/cosω) = -0.57/cos2ω

-(2 x 0.857) / (-0.57) = cosω/cos 2ω

3 = cosω/cos 2ω

3(cos 2ω) =  cosω

3(2cos²ω - 1) = cos ω

6cos²ω - 6 = cosω

6cos²ω  - cosω - 6 = 0

let cosω  = y

6y² - y - 6 = 0

solve the quadratic equation;

y = 1.1 or -0.92

cosω = -0.92

ω  = arc cos(-0.92)

ω  = 2.74 rad/s

From equation (1)

ln[-0.857 / cos(ω)] / x = -b  ---- (1)

let x = 1

ln(-0.857/cos(2.74) = -b

-0.93 = -b

b = 0.93

Thus, the damping coefficient is 0.93.

Learn more about damping coefficient here: brainly.com/question/14058210

#SPJ1

4 0
2 years ago
How many infrared photons of frequency 2.57 x 1013 Hz would need to be absorbed simultaneously by a tightly bound molecule to br
victus00 [196]

Answer:

94

Explanation:

f = 2.57 x 10^13 Hz

E = 10 eV = 10 x 1.6 x 10^-19 J = 1.6 x 10^-18 J

Energy of each photon = h f

Where, h is Plank's constant

Energy of each photon = 6.63 x 10^-34 x 2.57 x 10^13 = 1.7 x 10^-20 J

Number of photons = Total energy / energy of one photon

N = (1.6 x 10^-18) / (1.7 x 10^-20) = 94.11 = 94

6 0
3 years ago
If an electric wire is allowed to produce a magnetic field no larger than that of the Earth (0.503 X 104 T) at a distance of 15
dem82 [27]

Answer:

37.725 A

Explanation:

B = magnitude of the magnetic field produced by the electric wire = 0.503 x 10⁻⁴ T

r = distance from the wire where the magnetic field is noted = 15 cm = 0.15 m

i = magnitude of current flowing through the wire = ?

Magnetic field by a long wire is given as

B = \frac{\mu _{o}}{4\pi }\frac{2i}{r}

Inserting the values

0.503\times 10^{-4} = (10^{-7})\frac{2i}{0.15}

i = 37.725 A

6 0
3 years ago
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