A sound wave has a wavelength of 0.450 meters. If its speed in cold air is 330 meters/second, what is the wave's frequency?

2 answers:

7 0

Most of the information's required are already given in the question. Based on those information's the answer can be easily deduced.

Wavelength of the sound wave = 0.450 meters

Speed of the sound wave = 330 meters per second

We already know

v=fλ

330 = f * 0.450

f = 330/0.450

= 733.33 hertz

So the frequency of the wave is 733.33 hertz

faust18 [17]3 years ago

5 0

Most of the information's required are already given in the question. Based on those information's the answer can be easily deduced.
Wavelength of the sound wave = 0.450 meters
Speed of the sound wave = 330 meters per second
We already know
v=f<span>λ
</span>330 = f * 0.450
f = 330/0.450
= 733.33 hertz
So the frequency of the wave is 733.33 hertz

You might be interested in

Why does the frequency of a wave increase as the wavelength decreases

olchik [2.2K]

Explanation:

We know that the number of complete waves formed in 1 sec time is frequency and the distance between two consecutive crests or troughs is wavelength. And we have the formula that

Velocity = wavelength * frequency

or, frequency = velocity / wavelength

Here we can see frequency is directly proportional to velocity and indirectly proportional to wavelength.

So as the wavelength increases frequency decreases and as the wavelength decreases frequency increases.

Hope you understood

3 0

3 years ago

Read 2 more answers

You charge an initially uncharged 65.7-mf capacitor through a 39.1-Ï resistor by means of a 9.00-v battery having negligible int

uysha [10]

In a RC-circuit, with the capacitor initially uncharged, when we connect the battery to the circuit the charge on the capacitor starts to increase following the law:
$Q(t) = Q_0 (1-e^{-t/\tau})$
where t is the time, $Q_0 = CV$ is the maximum charge on the capacitor at voltage V, and $\tau = RC$ is the time constant of the circuit.
Using this law, we can answer all the three questions of the problem.

1) Using $R=39.1 \Omega$ and $C= 65.7 mF=65.7\cdot 10^{-3}F$ , the time constant of the circuit is:
$\tau = RC=(39.1 \Omega)(65.7 \cdot 10^{-3}F)=2.57 s$

2) To find the charge on the capacitor at time $t=1.95 \tau$ , we must find before the maximum charge on the capacitor, which is
$Q_0 = CV=(65.7 \cdot 10^{-3}F)(9 V)=0.59 C$
And then, the charge at time $t=1.95 \tau$ is equal to
$Q(1.95 \tau) = Q_0 (1-e^{-t/\tau})=(0.59 C)(1-e^{-1.95})=0.51 C$

3) After a long time (let's say much larger than the time constant of the circuit), the capacitor will be fully charged, this means its charge will be $Q_0 = 0.59 C$ . We can see this also from the previous formule, by using $t=\infty$ :
$Q(t) = Q_0 (1-e^{-\infty})=Q_0(1-0) = 0.59 C$

4 0

3 years ago

When a wheel is rotated through an angle of 21◦ , a point on the circumference travels through an arc length of 2.1 m. When the

LiRa [457]

Answer:

R = 5.73 m

Explanation:

For an angle of rotation through 21 degree we know that

arc length is given as

$angle \times Radius = Arc$