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melamori03 [73]
3 years ago
11

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 
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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>
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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]
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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
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