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ZanzabumX [31]
3 years ago
13

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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A 5000.0-kg car goes from 0 m/s to 70 m/s in 3600 seconds. What is the accelera
antiseptic1488 [7]

given:

mass = 5000 kg

u (initial velocity) = 0 m/s

v (final velocity) = 70 m/s

time taken to change velocity = 3600 s

acceleration = v - u / t

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a = 70 / 3600

a =  0.0194 m/s2 (approx)

given: mass = 5000 kg

acceleration = 0.0194 (found in 1st part)

force = mass * acceleration

f = ma

f = 5000*0.0194 = 97 N

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4 0
3 years ago
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bearhunter [10]

Explanation:

v = u + at

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a=-4

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7 0
3 years ago
Feathers and a bowling ball are dropped in a vacuum, airless environment. Which one will hit the ground first?
Reil [10]

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ΔP = 1.88 x 10^4 Pa. Use this answer to estimate the volume flow rate of blood from the head to the feet of a six-foot-tall pers
Sveta_85 [38]

Answer: 3765.66 \frac{m^{3}}{s}

Explanation:

We can solve this problem using the <u>Poiseuille equation</u>:

Q=\frac{\pi r^{4}\Delta P}{8\eta L}

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Q  is the Volume flow rate

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L=6 ft \frac{0.3048 m}{1 ft}=1.8288 m  is the length

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\eta=3(10)^{-3} Pa.s is the viscosity of blood

Solving:

Q=\frac{\pi (0.23 m)^{4}(1.88(10)^{4} Pa)}{8(3(10)^{-3} Pa.s)(1.8288 m)}

Q=3765.66 \frac{m^{3}}{s}

7 0
3 years ago
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