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2 years ago

What happens to the sound an object makes when the speed of vibrations decreases (slow down)?.

Physics

1 answer:

Masja [62]2 years ago

3 0

The frequency decreases, and you will hear a lower tone

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what is cut off from the open sea coral reefs or sandbars A watershed B lagoon C reef or D atoll . also this is a science questi

Klio2033 [76]

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3 years ago

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A 4 cm diameter "bobber" with a mass of 3 grams floats on a pond. A thin, light fishing line is tied to the bottom of the bobber

Tasya [4]

Answer:

Explanation:

Calculate the volume of the lead

$V=\frac{m}{d}\\\\=\frac{10g}{11.3g'cm^3}$

Now calculate the bouyant force acting on the lead

$F_L = Vpg$

$F_L=(\frac{10g}{11.3g/cm^3} )(1g/cm^3)(9.8m/s^2)\\\\=8.673\times 10^{-3}N$

This force will act in upward direction

Gravitational force on the lead due to its mass will act in downward direction

Hence the difference of this two force

$T=mg-F_L\\\\=(10\times10^{-3}kg(9.8m/s^2)-8.673\times 10^{-3}\\\\=8.933\times10^{-3}N$

If V is the volume submerged in the water then bouyant force on the bobber is

$F_B=V'pg$

Equate bouyant force with the tension and gravitational force

$F_B=T_mg\\\\V'pg=\frac{(8.933\times10^{-2}N)+mg}{pg} \\\\V'=\frac{(8.933\times10^{-2}N)+mg}{pg}$

Now Total volume of bobble is

$\frac{V'}{V^B} =\frac{\frac{(8.933\times10^{-2})+Mg}{pg} }{\frac{4}{3} \pi R^3 }\times100\\\\=\frac{\frac{(8.933\times10^{-2})+(3)(9.8)}{(1000)(9.8)} }{\frac{4}{3} \pi (4.0\times10^{-2})^3 }\times100\\\\$

= $\large\boxed{4.52 \%}$

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3 years ago

Electrons and protons are basic particles (and together make atoms) that have charge. What is the electric charge of an electron

trapecia [35]

Answer:

hi, here's the answer

Explanation:

the electric charge of an electron is negative (-ve).

pls mark this as the brainliest...

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3 years ago

Question 6 (5 points)

Ahat [919]

Answer:

its how life meets water and earth meets air.

Explanation:

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3 years ago

To practice Problem-Solving Strategy 16.1 Standing waves. An air-filled pipe is found to have successive harmonics at 800 HzHz ,

bazaltina [42]

Answer:

Length of the pipe = 53.125 cm

Explanation:

given data

harmonic frequency f1 = 800 Hz

harmonic frequency f2 = 1120 Hz

harmonic frequency f3 = 1440 Hz

solution

first we get here fundamental frequency that is express as

2F = f2 - f1 ...............1

put here value

2F = 1120 - 800

F = 160 Hz

and

Wavelength is express as

Wavelength = Speed ÷ Fundamental frequency ................2

here speed of waves in air = 340 m/s

so put here value

Wavelength =340 ÷ 160

Wavelength = 2.125 m

Length of the pipe will be

Length of the pipe = 0.25 × wavelength ......................3

put here value

Length of the pipe = 0.25 × 2.125

Length of the pipe = 0.53125 m

Length of the pipe = 53.125 cm

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2 years ago