Ask question

Ask question

All categories

valentina_108 [34]

3 years ago

8

A sound wave has a frequency of 500 Hz and a wavelength of 1.8 m. What is the wave speed of the sound wave? Question 1 options:

900 meters 900 m/s 278 meters 278 m/s

1 answer:

antoniya [11.8K]3 years ago

7 0

Answer:

The wave speed of the sound wave is 900 $\frac{m}{s}$ .

Explanation:

Wavelength is the minimum distance between two successive points on the wave that are in the same state of vibration. It is expressed in units of length (m).

Frequency is the number of vibrations that occur in a unit of time. Its unit is s⁻¹ or hertz (Hz).

The propagation velocity is the speed with which the wave propagates in the medium, that is, it is the magnitude that measures the speed at which the wave disturbance propagates along its displacement. Relate the wavelength (λ) and the frequency (f) inversely proportional using the following equation: v = f * λ.

In this case:

f= 500 Hz
λ= 1.8 m

Replacing:

v= 500 Hz* 1.8 m

v= 900 $\frac{m}{s}$

<u><em>The wave speed of the sound wave is 900 </em></u> $\frac{m}{s}$ <u><em>.</em></u>

You might be interested in

Explain why wine glasses with long stems have broad base

Anna35 [415]

Answer:

The best glasses have a wider bowl than rim to allow for proper swirling. The swirl releases volatile aroma compounds and creates a vortex in the center of the glass towards which these compounds are drawn

Explanation:\\\

6 0

3 years ago

How are sound waves different from the ripples that spread across water?

prisoha [69]

Answer:

.

Explanation:

Sound waves are caused by vibration ripples in water are caused by disturbance in the molecules

6 0

3 years ago

Read 2 more answers

A 1 200-kg car traveling initially at vCi 5 25.0 m/s in an easterly direction crashes into the back of a 9 000-kg truck moving i

sukhopar [10]

Answer:

The velocity of the truck after the collision is 20.93 m/s

Explanation:

It is given that,

Mass of car, m₁ = 1200 kg

Initial velocity of the car, $v_{Ci}=25\ m/s$

Mass of truck, m₂ = 9000 kg

Initial velocity of the truck, $v_{Ti}=20\ m/s$

After the collision, velocity of the car, $v_{Cf}=18\ m/s$

Let $v$ is the velocity of the truck immediately after the collision. The momentum of the system remains conversed.

$initial\ momentum=final\ momentum$

$1200\ kg\times 25\ m/s+9000\ kg\times 20\ m/s=1200\ kg\times 18+9000\ kg\times v$

$210000-21600=9000\ kg\times v$

$v=20.93\ m/s$

So, the velocity of the truck after the collision is 20.93 m/s. Hence, this is the required solution.

8 0

3 years ago

What is the displacement of the runner in 4 laps? *<br> 2 point

Serhud [2]

Answer:

There is no displacement.

Explanation:

Because the runner is running laps and returning to the original place, there is no displacement as displacement is relative to the change in location from the original position.

Hope this helps. . .

ly UwU

3 0

3 years ago

After flying for 15 min in a wind blowing 42 km/h at an angle of 19° south of east, an airplane pilot is over a town that is 48

masha68 [24]

Answer:

The speed of the airplane relative to the air is 209.47km/hr

Explanation:

Whenever we are solving a physics problem, it's really useful to start by drawing a diagram of the problem (See picture attached). It will help us visualize the problem better.

Now, we know that the plane flew for an amount of time of 15 minutes. For our dimensions to be the same, we need to turn those 15min to hours, like this:

$15min*\frac{1hr}{60min}=0.25hr$

Once our time is rewritten as hours, we can now calculate the velocity towards north of the plane.

$V=\frac{distance}{time}$

the plane traveled a distance to the north of 48km so the velocity is:

$V=\frac{48km}{0.25hr}$

so

V=192km/hr j

Now, we can calculate the x and y-components of the velocity of the wind. The problem states that the wind is blowing at 42km/hr at an angle of 19° south of east, so the x and y-components of the velocity of the wind are:

$V_{x}=42km/hr*cos(-19^{o} )=39.71 i$

and

$V_{y}=42km/hr*sin(-19^{o} )=-13.67 j$

So the velocity of the wind can be expressed as a vector as:

$V_{wind}=(39.71i - 13.67j)km/hr$

Once we know this, we can find the velocity of the plane with respect of the wind on x and on y:

$V_{plane x}=V_{plane/wind x}+V_{wind x}$

$V_{plane/wind x}=V_{plane x}-V_{wind x}$

$V_{plane/wind x}=(0-39.71 i)km/hr$

$V_{plane/wind x}= -39.71 i km/hr$

and

$V_{plane y}=V_{plane/wind y}+V_{wind y}$

$V_{plane/wind y}=V_{plane y}-V_{wind y}$

$V_{plane/wind y}=192km/hr j - (- 13.67j)km/hr$

$V_{plane/wind x}= 205.67 j km/hr$

So the velocity of the plane with respect to the wind can be rewritten as:

$V_{plane/wind x}= (-39.71i + 205.67 j) km/hr$

Since the problem asks us to find the speed of the plane with respect to the wind, this means that we need to find the magnitude of the velocity, since the speed is a scalar defined to be the magnitude of the velocity.

so:

$speed=\sqrt{(-39.71)^{2}+(205.67)^{2} }$

speed= 209.47 km/hr

Therefore, the speed of the airplane relative to the air is 209.47km/hr

6 0

3 years ago