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aliya0001 [1]
3 years ago
5

Lowest frequency of wave

Physics
2 answers:
Harlamova29_29 [7]3 years ago
4 0
The answer would be a radio wave
Nutka1998 [239]3 years ago
3 0
The lowest frequency portion of the electromagnetic spectrum is designated as “radio,” generally considered to have wavelengths within 1 millimeter to 100 kilometers or frequencies within 300 GHz to 3 kHz.
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A tuning fork vibrating at 508 Hz falls from rest and accelerates at 9.80 m/s^2. How far below the point of release is the tunin
JulijaS [17]

Answer:

Explanation:

given,

tuning fork vibration = 508 Hz

accelerates = 9.80 m/s²

speed of sound = 343 m/s

observed frequency = 490 Hz

f_s = f(\dfrac{v}{v-(-v_s)})

f_s = f(\dfrac{v}{v+v_s})

v_s = v[\dfrac{f_s}{f_o}-1]

      = 343[\dfrac{508}{490}-1]

      v_s=12.6 m/s

distance the tunning fork has fallen

y_1=\dfrac{v^2}{2a_y}

     =\dfrac{12.6^2}{2\times 9.8}

     =8.1 m

now, time required for the observed will be

t = \dfrac{8.1}{343} = 0.023 s

now, for the distance calculation

y_2 = u\ t + \dfrac{1}{2}at^2

  = 12.6\times 0.023 +\dfrac{1}{2}\times 9.8 \times 0.023^2

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total distance

 = 8.1 + 0.293 = 8.392 m

3 0
3 years ago
Question 2.
boyakko [2]

The tensile stress of the wire supporting 2 kg mass is determined as 6.1 x 10⁷ N/m².

<h3>Tensile stress of the wire</h3>

The tensile stress of the wire is calculated as follows;

σ = F/A

where;

  • A is area of the wire

A = πr² = πD²/4

where;

  • D is diameter = 0.64 mm

A = π x (0.64 x 10⁻³)²/4

A = 3.22 x 10⁻⁷ m²

σ = F/A = (mg)/A = (2 x 9.8)/( 3.22 x 10⁻⁷)

σ = 6.1 x 10⁷ N/m²

Learn  more about tensile stress here: brainly.com/question/25748369

#SPJ1

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