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3 years ago

If you increase the frequency of a sound wave four times what will happen to its speed

Physics

2 answers:

Anika [276]3 years ago

8 0

Answer: The correct answer is "the speed of the wave becomes four times".

Explanation:

The relation between the speed, frequency and the wavelength is as follows:

v=f\lambda

Here, v is the speed of the wave, f is the frequency and \lambda is the wavelength.

The speed of the sound wave is directly proportional to the frequency.

In the given problem, if the speed of the sound wave is increased four times then the speed of the sound becomes four times.

Therefore, the speed of the sound wave becomes four times.

Serggg [28]3 years ago

4 0

A w a v e ' s f r e q u e n c y c a n b e m a t h e m a t i c a l l y e x p r e s s e d a s f = v / λ, w h e r e v i s t h e s p e e d a n d λ i s t h e w a v e l e n g t h . W e c a n s e e t h a t f r e q u e n c y a n d s p e e d a r e d i r e c t l y p r o p o r t i o n a l to e a c h o t h e r , t h a t m e a n s w h e n f r e q u e n c y i s i n c r e a s e d f o u r t i m e s, t h e s p e e d w i l l a l s o i n c r e a s e f o u r t i m e s<span>.</span>

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3 years ago

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2 years ago

The storm surge of a hurricane can cause

BabaBlast [244]

Answer:

it can cause huricans and flooding

Explanation:

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3 years ago

Read 2 more answers

A 1.5m long string weighs 0.0020 kg. It is tensioned to 100N. A disturbance travels along it with a wavelength of 1.5m, find:a)

Zigmanuir [339]

Answer:

the propagation velocity of the wave is 274.2 m/s

Explanation:

Given;

length of the string, L = 1.5 m

mass of the string, m = 0.002 kg

Tension of the string, T = 100 N

wavelength, λ = 1.5 m

The propagation velocity of the wave is calculated as;

$v = \sqrt{\frac{T}{\mu} } \\\\\mu \ is \ mass \ per \ unit \ length \ of \ the \ string\\\\\mu = \frac{0.002 \ kg}{1.5 \ m} = 0.00133 \ kg/m\\\\v = \sqrt{\frac{100}{0.00133} } \\\\v = 274.2 \ m/s$

Therefore, the propagation velocity of the wave is 274.2 m/s

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3 years ago

A proton and an electron are fixed in space with a separation of 859 nm. Calculate the electric potential at the midpoint betwee

makkiz [27]

Answer:

The electric potential at the midpoint between the two particles is 3.349 X 10⁻³ Volts

Explanation:

Electric potential is given as;

V = E*r

where;

E is the electric field strength, = kq/r²

V = ( kq/r²)*r

V = kq/r

k is coulomb's constant = 8.99 X 10⁹ Nm²/C²

q is the charge of the particles = 1.6 X 10⁻¹⁹ C

r is the distance between the particles = 859 nm

At midpoint, the distance = r/2 = 859nm/2 = 429.5 nm

V = (8.99 X 10⁹ * 1.6 X 10⁻¹⁹)/ (429.5 X 10⁻⁹)

V = 3.349 X 10⁻³ Volts

Therefore, the electric potential at the midpoint between the two particles is 3.349 X 10⁻³ Volts

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3 years ago