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Ann [662]
3 years ago
12

Two loudspeakers (A and B) are 3.20m apart and emitting a sound with a frequency of 400Hz. An observer is 2.10m directly in fron

t of A. If the speed of sound in this room is 340m/s will the observer hear a loud sound or a quiet sound?
Physics
1 answer:
TiliK225 [7]3 years ago
5 0

Answer:

The observer hears a loud sound

Explanation:

In order to know if the observer hears a loud or a quiet sound, you need to know if there is a constructive or destructive interference between the sound waves of the loudspeakers.

You first calculate the distance between the observer and the loudspeakers.

The distances are given by:

d1: distance to loudspeaker A = 2.10m

d2: distance to loudspeaker B

d_2=\sqrt{(3.20m)^2+(2.10m)^2}=3.827m

Next, you calculate the wavelength of the sound waves by using the following formula:

\lambda=\frac{v_s}{f}

vs: speed of sound =  343 m/s

f: frequency of the waves = 400Hz

λ: wavelength

\lambda=\frac{343m/s}{400Hz}=0.8575m

Next, you calculate the path difference between the distance from the observer to the loudspeakers:

\Delta d=3.827m-2.10m=1.727m

You obtain a constructive interference (loud sound) if the quotient between the wavelength of the sound and the difference path is an integer:

\frac{\Delta d}{\lambda}=\frac{1.727m}{0.857}\approx2

Then, there will be a constructive interference, and the sound who the observer hears is loud.

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6 0
1 year ago
A truck tire rotates at an initial angular speed of 21.5 rad/s. The driver steadily accelerates, and after 3.50 s the tire's ang
erma4kov [3.2K]

Given:

initial angular speed, \omega _{i} = 21.5 rad/s

final angular speed, \omega _{f} = 28.0 rad/s

time, t = 3.50 s

Solution:

Angular acceleration can be defined as the time rate of change of angular velocity and is given by:

\alpha = \frac{\omega_{f} - \omega _{i}}{t}

Now, putting the given values in the above formula:

\alpha = \frac{28.0 - 21.5}{3.50}

\alpha = 1.86 m/s^{2}

Therefore, angular acceleration is:

\alpha = 1.86 m/s^{2}

5 0
3 years ago
A ball is dropped from the top of a 77 m building. With what speed does the ball hit the ground? _________ m/s
vitfil [10]

Answer:

38.87 m/s

Explanation:

Given that the ball is dropped from a height = 77 m

u = 0 m/s

s = 77 m

a = g = 9.81 m/s²

Applying the expression as:

v^2-u^2=2as

Applying values as:

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 77+0^2}\\\Rightarrow v=38.87\ m/s

<u>The speed with which the ball hit the ground = 38.87 m/s</u>

3 0
3 years ago
A professional baseball player can throw the ball around 45 m/s if the distance between the pitcher and the batter is 18.39 m. H
svetlana [45]

Answer:

The time taken for the ball to get to the batter is 0.41 s.

Explanation:

Given;

initial velocity of the baseball, u = 45 m/s

horizontal distance between the pitcher and the batter, X = 18.39 m

The horizontal distance or range of a projectile is given as;

X = ut

where;

t is the time of flight

u is the initial velocity

t = X / u

t = 18.39 / 45

t = 0.41 s

Therefore, the time taken for the ball to get to the batter is 0.41 s.

6 0
2 years ago
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