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PilotLPTM [1.2K]
3 years ago
6

If the speed of the wave is 12 m/s and the frequency is 2.3 Hz, what is the wavelength?

Physics
1 answer:
-BARSIC- [3]3 years ago
3 0

Answer: 5.2

Explanation: 12/ 2.3

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3 years ago
the density of aluminum is 2700 kg/m3. if transverse waves travel at in an aluminum wire of diameter what is the tension on the
Sunny_sXe [5.5K]

The tension on the wire is 52.02 N.

From the question, we have

Density of aluminum = 2700 kg/m3

Area,

A = πd²/4

A = π x (4.6 x 10⁻³)²/4

A = 1.66 x 10⁻⁵ m²

μ = Mass per unit length of the wire

μ = ρA

μ = 2700 kg/m³ x 1.66 x 10⁻⁵ m²

μ = 0.045 kg/m

Tension on the wire = √T/μ

34 = √T/0.045

34² = T/0.045

T = 52.02 N

The tension on the wire is 52.02 N.

Complete question:

The density of aluminum is 2700 kg/m3. If transverse waves propagate at 34 m/s in a 4.6-mm diameter aluminum wire, what is the tension on the wire.

To learn more about tension visit:  brainly.com/question/14336853

#SPJ4

5 0
1 year ago
Determine the angle between the directions of vector A with rightwards arrow on top = 3.00i + 1.00j and vector B with rightwards
sertanlavr [38]

Answer:

C) 26.6

Explanation:

I don't know how to calculate vector

6 0
3 years ago
An electron and a proton are held on an x axis, with the electron at x = + 1.000 m
mixas84 [53]

Answer:

  r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

Explanation:

For this exercise we must use conservation of energy

the electric potential energy is

          U = k \frac{q_1q_2}{r_{12}}

for the proton at x = -1 m

          U₁ =- k \frac{e^2 }{r+1}

for the electron at x = 1 m

          U₂ = k \frac{e^2 }{r-1}

starting point.

        Em₀ = K + U₁ + U₂

        Em₀ = \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1}

final point

         Em_f = k e^2 ( -\frac{1}{r_2 +1} + \frac{1}{r_2 -1})

   

energy is conserved

        Em₀ = Em_f

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e^2 (- \frac{1}{r_2 +1} + \frac{1}{r_2 -1})              

       

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e²(  \frac{2}{(r_2+1)(r_2-1)} )

we substitute the values

½ 9.1 10⁻³¹ 450 + 9 10⁹ (1.6 10⁻¹⁹)² [ - \frac{1}{20+1} + \frac{1}{20-1} ) = 9 109 (1.6 10-19) ²( \frac{2}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 2.304 10⁻³⁷ (5.0125 10⁻³) = 4.608 10⁻³⁷ ( \frac{1}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 1.1549 10⁻³⁹ = 4.608 10⁻³⁷     \frac{1}{r_2^2 -1}

          \frac{2.0475 \ 10^{-28} }{1.1549 \ 10^{-37} } = \frac{1}{r_2^2 -1}

          r₂² -1 = (4.443 10⁸)⁻¹

           

          r2 = \sqrt{1 + 2.25 10^{-9}}

          r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

4 0
3 years ago
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