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3 years ago

A 2.00 m wave has a frequency of 686 Hz. What is the speed of the wave?

Physics

1 answer:

qwelly [4]3 years ago

4 0

Wave speed = frequency * wavelength
Wave speed = 686 * 2.00
Wave speed = 1,372 (m/s)

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A sound wave is determined to have a frequency of 1,000 Hz and a wavelength of 35 cm. What is the speed of this wave? If the sou

mars1129 [50]

Answer:

350 m/s

31°C

Explanation:

Speed of sound is given as the product of frequency and wavelength

S=fw

Where s represent the speed in m/s, f is frequency and w is wavelength

Conversion

Taking 1m to be 100 cm then 35 cm will be 35/100=0.35m

Substituting 0.35 m for w and 1000 Hz for f then

S=1000*0.35=350 m/s

speed of sound (m/s) = 331.5 + 0.60 T(°C)

350=331.5+0.6T

T=30.833333333333

The temperature is approximately 31°C

4 0

3 years ago

What is the frequency of a microwave with a wavelength of 3.52 mm?

AlekseyPX

As per the question the wavelength of the microwave is given as 3.52 mm.

we are asked to calculate the frequency of the wave.

we know that microwave is a electromagnetic wave.

As per Clark Maxwell's electromagnetic theory ,every electromagnetic wave moves with a velocity equal to the velocity of light in vacuum and that is equal to 3×10^8 m/s.

From the equation of the wave,we know that velocity of wave is the product of frequency and wavelength.

Mathematically wave velocity $v=f*\lambda$ where f is the frequency of the wave and $\lambda$ is the wavelength.

As per the question $\lambda=3.52 mm$

$= 3.52*10^{-3} m$

Here $v=c=3*10^{8} m/s$

Hence frequency of the wave $f=\frac{v}{\lambda}$

$=\frac{3*10^{8} }{3.52*10^{-3} } s^{-1}$

$=0.852272727272*10^{11} Hz$

Here Hertz [Hz] is the unit of frequency.

6 0

3 years ago

Read 2 more answers

VMariaS [17]

Answer:

no they cannot, only opposite charges attract

5 0

3 years ago

Read 2 more answers

A flat surface is in a uniform magnetic field. Given only the area of the surface and the magnetic flux through the surface, it

Tasya [4]

Answer:

Given the area A of a flat surface and the magnetic flux through the surface $\Phi$ it is possible to calculate the magnitude $\frac{\Phi}{A}=B\ cos \theta$ .

Explanation:

The magnetic flux gives an idea of how many magnetic field lines are passing through a surface. The SI unit of the magnetic flux $\Phi$ is the weber (Wb), of the magnetic field B is the tesla (T) and of the area A is ( $m^{2}$ ). So 1 Wb=1 T.m².

For a flat surface S of area A in a uniform magnetic field B, with $\theta$ being the angle between the vector normal to the surface S and the direction of the magnetic field B, we define the magnetic flux through the surface as: