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andrew11 [14]
3 years ago
10

If the frequencies of two component waves are 24 Hz and 20 Hz, they should produce _______ beats per second.

Physics
2 answers:
horrorfan [7]3 years ago
8 0
This can be answered using the beat frequency formula, which is simply the difference between 2 frequencies.

Let: <span>fᵇ = beat frequency
</span>f₁ = first frequency
f₂ = second frequency

fᵇ = |f₁ - f₂|

substituting the values:
fᵇ = |24Hz - 20Hz|
fᵇ = 4Hz

The unit Hz also means beats per second, therefore:
<span>fᵇ = 4 beats per second
</span>
Therefore, the answer is C. 4
mrs_skeptik [129]3 years ago
3 0

C. 4 beats per second is correct

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Answer:

a)32.34 N/m

b)10cm

c)1.6 Hz

Explanation:

Let 'k' represent spring constant

'm' mass of the object= 330g =>0.33kg

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ΣF=kx-mg=0

k=mg / x

k= (0.33 x 9.8)/ 0.1

k= 32.34 N/m

b) The amplitude, A, is the distance from the equilibrium (or center) point of motion to either its lowest or highest point (end points). The amplitude, therefore, is half of the total distance covered by the oscillating object.

Therefore, amplitude of the oscillation is 10cm

c)frequency of the oscillation can be determined by,

f= 1/2π \sqrt{\frac{k}{m} }

f= 1/2π \sqrt{\frac{32.34}{0.33} }

f= 1.57

f≈ 1.6 Hz

Therefore,  the frequency of the oscillation is 1.6 Hz

5 0
3 years ago
A physics student looks into a microscope and observes that small particles suspended in water are moving about in an irregular
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Answer:

d. the actual motion is regular, but the speeds of particles are too large to observe the regular motion

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What do we call the principal SI units that are used to derive all other SI units? Question 3 options:
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5 0
2 years ago
A section of a parallel-plate air waveguide with a plate separation of 7.11 mm is constructed to be used at 15 GHz as an evanesc
adell [148]

Answer:

the required minimum length of the attenuator is 3.71 cm

Explanation:

Given the data in the question;

we know that;

f_{c_1 = c / 2a

where f is frequency, c is the speed of light in air and a is the plate separation distance.

we know that speed of light c = 3 × 10⁸ m/s = 3 × 10¹⁰ cm/s

plate separation distance a = 7.11 mm = 0.0711 cm

so we substitute

f_{c_1 = 3 × 10¹⁰ / 2( 0.0711  )

f_{c_1 = 3 × 10¹⁰ cm/s / 0.1422  cm

f_{c_1 =  21.1 GHz which is larger than 15 GHz { TEM mode is only propagated along the wavelength }

Now, we determine the minimum wavelength required

Each non propagating mode is attenuated by at least 100 dB at 15 GHz

so

Attenuation constant TE₁ and TM₁ expression is;

∝₁ = 2πf/c × √( (f_{c_1 / f)² - 1 )

so we substitute

∝₁ = ((2π × 15)/3 × 10⁸ m/s) × √( (21.1 / 15)² - 1 )

∝₁ = 3.1079 × 10⁻⁷

∝₁ = 310.79 np/m

Now, To find the minimum wavelength, lets consider the design constraint;

20log₁₀e^{-\alpha _1l_{min = -100dB

we substitute

20log₁₀e^{-(310.7np/m)l_{min = -100dB

l_{min = 3.71 cm

Therefore, the required minimum length of the attenuator is 3.71 cm

6 0
2 years ago
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