All categories

3 years ago

What is a wavelength of sound wave moving at 340m/s with frequency of 256hz

Physics

1 answer:

zimovet [89]3 years ago

5 0

V = 340 m/s
f = 256 Hz
lambda (wavelength)

v = f*lambda
340 = 256 * lambda
340/256 = lambda
lambda = 1.328 m

You might be interested in

What element x is most likely to react to form the compound xf5?

djyliett [7]

<span>antimony. It has +3,+5,-3 so yeah. the others carbon+2,+4,-4, chlorine +1,+5,+7,-1 but -1 is the most often so it isn't Cl, calcium +2.</span>

8 0

3 years ago

A toy cannon launches a 46-g golf ball straight up into the air with a kinetic energy of 6.8 J. What must the

Shkiper50 [21]

Answer : The correct option is, (C) 17 m/s

Explanation :

Formula used :

$K.E=\frac{1}{2}mv^2$

where,

K.E = kinetic energy = 6.8 J

m = mass of object = 46 g = 0.046 kg (1 kg = 1000 g)

v = velocity

Now put all the given values in the above formula, we get:

$K.E=\frac{1}{2}mv^2$

$6.8J=\frac{1}{2}\times 0.046kg\times v^2$

$6.8kg.m^2/s^2=\frac{1}{2}\times 0.046kg\times v^2$

$v=17.19m/s\approx 17m/s$

Therefore, the ball's velocity be as it leaves the cannon is, 17 m/s

3 0

3 years ago

Which statements about velocity are true?

OleMash [197]

<span>the speed of a direction</span>

5 0

3 years ago

Read 2 more answers

A listener increases his distance from a sound source by a factor of 4.49.

noname [10]

Answer: Δβ (dB) = -13.1dB

Explanation:

The intensity of sound is inversely proportional to the square of the distance between them.

I ∝ 1/r²

I₁/I₂= r₂²/r₁² .....1

When the listener increases his distance from the source by a factor of 4.49.

Then,

r₂/r₁= 4.49

From equation 1

I₁/I₂ = (4.49)²

I₁/I₂ = 20.16

I₂/I₁ = 1/20.16

The change in sound intensity in dB can be given as

Δβ (dB) = 10 log(I₂/l₁) = 10log(1/20.6) = -13.1dB

6 0

3 years ago

Light of wavelength 648.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 57.5 cm from the slit.

boyakko [2]

Answer:

$71.0 \mu m$

Explanation:

The formula for the single-slit diffraction is

$y=\frac{n\lambda D}{d}$

where

y is the distance of the n-minimum from the centre of the diffraction pattern

D is the distance of the screen from the slit

d is the width of the slit

$\lambda$ is the wavelength of the light

In this problem,

$\lambda=648.0 nm=6.48\cdot 10^{-7}m$

$D=57.5 cm=0.575 m$

$y=1.05 cm=0.0105 m$ , with n=2 (this is the distance of the 2nd-order minimum from the central maximum)

Solving the formula for d, we find:

$d=\frac{n\lambda D}{y}=\frac{2(6.48\cdot 10^{-7} m)(0.575 m)}{0.0105 m}=7.10\cdot 10^{-5} m= 71.0 \mu m$

6 0

3 years ago