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Afina-wow [57]
3 years ago
8

If the third harmonic produced by a musical instrument is 300 HZ, then the first overtone will be?​

Physics
1 answer:
SCORPION-xisa [38]3 years ago
7 0

Answer:

the first overtone is 200 Hz

Explanation:

The computation of the first overtone would be given below:

The following formula should be applied

n_1 = n_3 ÷ 3

= 300 ÷ 3

= 100 Hz

Now n_2 is

= 2 × 100

= 200 Hz.

Hence, the first overtone is 200 Hz

we simply applied the above formula and the same would be relevant

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A. attract each other.

The Law of Universal Gravitation discusses the phenomenon of gravity. Remember that gravity is the force that keeps us on Earth; the Earth pulls us down, and our bodies pull back.  Gravity is the force of attraction, so the correct answer is a).
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3 years ago
Calculate the change in length of a 90.5 mm aluminum bar that has increased in temperature by from -14.4 oC to 154.6 oC
nignag [31]

Answer:

 ΔL = 3.82 10⁻⁴ m

Explanation:

This is a thermal expansion exercise

          ΔL = α L₀ ΔT

          ΔT = T_f - T₀

where ΔL is the change in length and ΔT is the change in temperature

Let's reduce the length to SI units

          L₀ = 90.5 mm (1m / 1000 mm) = 0.0905 m

let's calculate

          ΔL = 25.10⁻⁶ 0.0905 (154.6 - (14.4))

          ΔL = 3.8236 10⁻⁴ m

     

using the criterion of three significant figures

          ΔL = 3.82 10⁻⁴ m

5 0
3 years ago
Which is an si metric unit of measurement that is used to record the heat transfer of a solution in a classroom investigation?
kumpel [21]
The SI unit for heat energy is joule
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When energy decreases, particle motion:<br> A. stops<br> B. increases<br> C. decreases
DaniilM [7]
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Read 2 more answers
A two-slit pattern is viewed on a screen 1.26 m from the slits. If the two fourth-order maxima are 53.6 cm apart, what is the to
Anettt [7]

Answer:

= 6.55cm

Explanation:

Given that,

distance = 1.26 m

distance between  two fourth-order maxima = 53.6 cm

distance between central bright fringe and fourth order maxima

y = Y / 2

  =  53.6cm / 2

  = 26.8 cm

  =0.268 m

tan θ = y / d

         = 0.268 m /  1.26 m

         = 0.2127

       θ = 12°

4th maxima

d sinθ = 4λ

d / λ = 4 / sinθ

d / λ = 4 / sin 12°

d / λ = 19.239

for first (minimum)

d sinθ = λ / 2

sinθ =  λ / 2d

       =  1 / 2(19.239)

       = 1 / 38.478

       = 0.02599

    θ =  1.489°

tan θ = y / d

y = d tan θ

  = 1.26 tan 1.489°

  = 0.03275

the total width of the central bright fringe  

Y = 2y

  = 2(0.03275)

  = 0.0655m

  = 6.55cm

4 0
3 years ago
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