All categories

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>

You might be interested in

A girl weighing 600 N steps on a bathroom scale that contains a stiff spring. In equilibrium, the spring is compressed 1.0 cm un

8_murik_8 [283]

Answer:

The spring constant is 60,000 N

The total work done on it during the compression is 3 J

Explanation:

Given;

weight of the girl, W = 600 N

compression of the spring, x = 1 cm = 0.01 m

To determine the spring constant, we apply hook's law;

F = kx

where;

F is applied force or weight on the spring

k is the spring constant

x is the compression of the spring

k = F / x

k = 600 / 0.01

k = 60,000 N

The total work done on the spring = elastic potential energy of the spring, U;

U = ¹/₂kx²

U = ¹/₂(60000)(0.01)²

U = 3 J

Thus, the total work done on it during the compression is 3 J

3 0

3 years ago

Two balls have the same mass of 5.00 kg. Suppose that these two balls are attached to a rigid massless rod of length 2L, where L

olga_2 [115]

Answer:

Explanation:

Given

mass of balls $m= 5 kg$

$N=45.6 rev/s$

angular velocity $\omega =2\pi N=286.55 rad/s$

Length of Rod $2L=1.1 m$

Tension in the Second half of rod

$T_2=m\omega ^2(2L)=2m\omega ^2L$

$T_2=5\times (286.55)^2\times 1.1$

$T_2=451.609 kN$

For First Part

$T_1-T_2=m\omega ^2L$

$T_1=T_2+m\omega ^2L$

$T_1=3 m\omega ^2L$

$T_1=3\times 5\times (286.55)^2\times 0.55$

$T_1=677.41 kN$

7 0

3 years ago

a 5.00 × 105 kg rocket is accelerating straight up. Its engines produce 1.50 × 107 of thrust, and air resistance is 4.50 × 106 N

Angelina_Jolie [31]

This one is simple :)

use the equation F = ma and then re-arrange for a.

a = F / m

But there is a trick, so be careful. the question gives you wind resistance. Simply subtract the wind resistance from the thrust of the rocket to get the net force upward.

$1.50 * 10^7N - 4.5 8 10^6N = 1.05 * 10^7N$

So,

$a = \frac{F}{m} = \frac{1.05*10^7N}{5.00*10^5kg}$