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3 years ago

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Waves must exhibit a constant phase difference for interference to occur. Please select the best answer from the choices provide

d T F

1 answer:

zalisa [80]3 years ago

8 0

True is what it is/ ,,,,,,,,,,,,,

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What happens to parallel light rays that strike a concave lens?

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Due to the shape of the lens , parallel rays will be deviated

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Two loudspeakers placed 8.0 m apart are driven in phase by an audio oscillator whose frequency range is 2.2 kHz to 2.9 kHz. A po

My name is Ann [436]

Answer:

The answer to the question is 2.2khz

Explanation:

<em>Let z₁ = 5.4m</em>

<em>Let z₂ = 4.6m</em>

<em>The path difference Δz = z₁-z₂ = 5.4 - 4.6 = 0.8m</em>

<em>For the interference= Δz λ, 2λ, 3λ......</em>

<em>The wavelength λ = 0.8m</em>

<em>The speed of sound v = 344m/s</em>

<em>The frequency f = v/λ = 344/0.8 = 430hz</em>

<em>Now,</em>

<em>f₁ =f, f₂= 2f, f₃ = 3f, f₄= 4f, f₅ =5f which is,</em>

<em>f₁ =f = 430Hz, f₂=2f =860Hz, f₃ =3f =1290Hz f₄ =4f =1720Hz and f₅=5f =2150Hz</em>

<em>f5 = 2120Hz = 2.200Hz </em>

<em>we will convert to two significant figures =2.2kHz</em>

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3 years ago

A string with a mass density of 3 * 10^-3 kg/m is under a tension of 380 N and is fixed at both ends. One of its resonance frequ

Delvig [45]

Answer:

(a) the fundamental frequency of this string is 65 Hz

(b) the harmonics of the given frequencies are third and fourth respectively.

(c) the length of the string is 2.74 m

Explanation:

Given;

mass density of the string, μ = 3 x 10⁻³ kg/m

tension of the string, T = 380 N

resonating frequencies, 195 Hz and 260 N

For the given resonant frequencies;

$195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } ---(1)\\\\260 = \frac{n+1}{2l} \sqrt{\frac{T}{\mu} } ---(2)\\\\divide \ (2) \ by (1)\\\\\frac{260}{195} = \frac{n+1 }{n} \\\\260n = 195(n+1)\\\\260 n = 195 n + 195\\\\260n - 195n = 195\\\\65n = 195\\\\n = \frac{195}{65} \\\\n = 3$

(c) From any of the equations, solve for Length of the string (L);

$195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } \\\\195 = \frac{3}{2l}\sqrt{\frac{380}{3\times 10^{-3}} } \\\\l = \frac{3}{2\times 195}\sqrt{\frac{380}{3\times 10^{-3}} }\\\\l = 2.74 \ m$

(a) the fundamental frequency is calculated as;

$f_o = \frac{1}{2l} \sqrt{\frac{T}{\mu} } \\\\f_o = \frac{1}{2\times 2.74} \sqrt{\frac{380}{3\times 10^{-3} } }\\\\f_o = 65 \ Hz$

(b) harmonics of the given frequencies;

the first harmonic (n = 1) = f₀ = 65 Hz

the second harmonic (n = 2) = 2f₀ = 130 Hz

the third harmonic (n = 3) = 3f₀ = 195 Hz

the fourth harmonic (n = 4) = 4f₀ = 260 Hz

Thus, the harmonics of the given frequencies are third and fourth respectively.

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Which statement does not describe the relationship between genes and

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Answer: C. Inherited traits carry the instructions for individual genes.

Explanation: bejewels I know stuff. ☆ - ~ hope this helps

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3 years ago

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In making the determination that bullets, shells, or cartridge cases were fired from a specific weapon, the criminalist would us

ss7ja [257]

Answer:

Comparison Microscope

Explanation:

The Comparison Microscope allows for comparison between two objects or samples by placing them side by side.

It is primarily used in criminology for ballistics which makes it ideal to find out if bullets, shells, or cartridge cases were fired from a specific weapon.

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3 years ago