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kogti [31]
3 years ago
10

I NEED a LOT of help PLEASE!

Physics
2 answers:
Minchanka [31]3 years ago
3 0
Infrared, visible light, then ultraviolet. Infrared is light that the human eye can not see and visible light is clearly light we can see then ultraviolet is has such a high frequency we can't see it either.
KIM [24]3 years ago
3 0
It is B. Infrared, visible, ultraviolet.
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7 hectometers is 7000 decimeters

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When waves interfere together and result in a quieter sound is called
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That is called fully constructive interference.
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What's the differentiating characteristic between an element and a compound
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an element is a pure substance whereas a compound is a mixture of two or more elements.

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Read 2 more answers
Assume a form y1 = A sin(α) for the transverse displacement of the string. Enter an expression for α of a transverse wave on a s
maria [59]

Answer:

α = 2Πft + 2Πx/¶

Explanation:

A transverse wave is one of the type of electromagnetic waves which occurs when the vibration of the medium is perpendicular to the direction of the wave. Mathematically, the wave equation according to the question

y1 = Asin(α).... (1)

since its characteristic wave gives a sine wave.

y1 is the vertical displacement

A is the amplitude.

The expression for α of the transverse wave on a string traveling along the positive x-direction in terms of its wavenumber k, the position x, its angular frequency ω, and the time t can be given as;

α = ωt+theta ... (2)

theta is the phase angle. This phase angle is a function of the position x of the wave and the wave number.

theta = kx

where k = 2Πx/¶... (3)

¶ is the wavelength.

ω which is the angular frequency is a function of the frequency (f) of the wave and it is given as;

ω = 2Πf... (4)

Substituting equation 3 and 5 into 2;

α = 2Πft + 2Πx/¶

Therefore,

y1 = Asin(2Πft + 2Πx/¶)

4 0
3 years ago
You have a stopped pipe of adjustable length close to a taut 85.0cm, 7.25g wire under a tension of 4170N. You want to adjust the
Makovka662 [10]

Answer:

The length is L_d= 0.069 \ m

Explanation:

From the question we are told that

               The length of the wire  L = 85cm = \frac{85}{100}  = 0.85m

                The mass is  m = 7.25g = \frac{7.25}{1000}  = 7.25^10^{-3}kg

                The tension is  T = 4170N

Generally the frequency of  oscillation of a stretched wire is mathematically represented as

             f = \frac{n}{2L} \sqrt{\frac{T}{\mu}

Where n is the the number of nodes = 3 (i.e the third harmonic)

             \mu is the linear mass density of the wire

 This linear mass density is mathematically represented as

               \mu = \frac{m}{L}

Substituting values

            \mu = \frac{7.2*10^{-3}}{0.85}

                = 8.53 *10^{-3} kg/m

 Substituting values in to the equation for frequency

            f = \frac{3}{2 80.85} * \sqrt{\frac{4170}{8.53*10^{-3}} }

               = 1234Hz

From the question the we can deduce that the fundamental frequency is equal to the oscillation of a stretched wire

The fundamental frequency is mathematically represented as

        f = \frac{v}{4L_d}

Where L_d is the  length of the pipe

          v is the speed of sound with a value of v = 343m/s

    Making  L_d the subject of the formula

                      L_d = \frac{v}{4f}

   Substituting values

                   L_d = \frac{343}{(4)(1234)}

                        L_d= 0.069 \ m

From the question the we can deduce that the fundamental frequency is equal to the oscillation of a stretched wire

6 0
3 years ago
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