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Elanso [62]
2 years ago
14

Provided the amplitude is sufficiently great, the human ear can respond to longitudinal waves over a range of frequencies from a

bout 20.0 Hz to about 20.0 kHz.a)If you were to mark the beginning of each complete wave pattern with a red dot for the long-wavelength sound, how far apart would the red dots be?b)If you were to mark the beginning of each complete wave pattern with a blue dot for the short-wavelength sound, how far apart would the blue dots be?c)In reality would adjacent red dots be far enough apart for you to easily measure their separation with a meterstick?d)In reality would adjacent blue dots be far enough apart for you to easily measure their separation with a meterstick?e)Suppose you repeated part A in water, where sound travels at 1480 {\rm{ m/s}}. How far apart would the red dots be ?f)Suppose you repeated part A in water, where sound travels at 1480 {\rm{ m/s}}. How far apart would the blue dots be ?g)Could you readily measure their separation with a meterstick?
Physics
1 answer:
riadik2000 [5.3K]2 years ago
8 0

Answer:

Check the explanation

Explanation:

The beat frequency is

df = f2 - f1

the wavelength is

lamda1 = (v/f1)

and lamda2 = (v/f2)

where v = 340 m/s,f1 = 25.0 kHz and f2 = 20.0 kHz

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Kenya has a pile of salt on the counter. She uses a pinch of it for a meal. Which of the following physical properties of the sa
SashulF [63]
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3 years ago
Read 2 more answers
A 217 Ω resistor, a 0.875 H inductor, and a 6.75 μF capacitor are connected in series across a voltage source that has voltage a
Nataly [62]

For an AC circuit:

I = V/Z

V = AC source voltage, I = total AC current, Z = total impedance

Note: We will be dealing with impedances which take on complex values where j is the square root of -1. All phasor angles are given in radians.

For a resistor R, inductor L, and capacitor C, their impedances are given by:

Z_{R} = R

R = resistance

Z_{L} = jωL

ω = voltage source angular frequency, L = inductance

Z_{C} = -j/(ωC)

ω = voltage source angular frequency, C = capacitance

Given values:

R = 217Ω, L = 0.875H, C = 6.75×10⁻⁶F, ω = 220rad/s

Plug in and calculate the impedances:

Z_{R} = 217Ω

Z_{L} = j(220)(0.875) = j192.5Ω

Z_{C} = -j/(220×6.75×10⁻⁶) = -j673.4Ω

Add up the impedances to get the total impedance Z, then convert Z to polar form:

Z = Z_{R} + Z_{L} + Z_{C}

Z = 217 + j192.5 - j673.4

Z = (217-j480.9)Ω

Z = (527.6∠-1.147)Ω

Back to I = V/Z

Given values:

V = (30.0∠0+220t)V (assume 0 initial phase, and t = time)

Z = (527.6∠-1.147)Ω (from previous computation)

Plug in and solve for I:

I = (30.0∠0+220t)/(527.6∠-1.147)

I = (0.0569∠1.147+220t)A

To get the voltages of each individual component, we'll just multiply I and each of their impedances:

v_{R} = I×Z_{R}

v_{L} = I×Z_{L}

v_{C} = I×Z_{C}

Given values:

I = (0.0569∠1.147+220t)A

Z_{R} = 217Ω = (217∠0)Ω

Z_{L} = j192.5Ω = (192.5∠π/2)Ω

Z_{C} = -j673.4Ω = (673.4∠-π/2)Ω

Plug in and calculate each component's voltage:

v_{R} = (0.0569∠1.147+220t)(217∠0) = (12.35∠1.147+220t)V

v_{L} = (0.0569∠1.147+220t)(192.5∠π/2) = (10.95∠2.718+220t)V

v_{C} = (0.0569∠1.147+220t)(673.4∠-π/2) = (38.32∠-0.4238+220t)V

Now we have the total and individual voltages as functions of time:

V = (30.0∠0+220t)V

v_{R} = (12.35∠1.147+220t)V

v_{L} = (10.95∠2.718+220t)V

v_{C} = (38.32∠-0.4238+220t)V

Plug in t = 22.0×10⁻³s into these values and take the real component (amplitude multiplied by the cosine of the phase) to determine the real voltage values at this point in time:

V = 30.0cos(0+220(22.0×10⁻³)) = 3.82V

v_{R} = 12.35cos(1.147+220(22.0×10⁻³)) = 11.8V

v_{L} = 10.95cos(2.718+220(22.0×10⁻³)) = 3.19V

v_{C} = 38.32cos(-0.4238+220(22.0×10⁻³)) = -11.2V

4 0
2 years ago
A 6,000N is applied to a formula one car that weighs 500kg. What is the car's acceleration?
Vesna [10]

Answer:

<h2>12 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

f is the force

m is the mass

From the question

f = 6000 N

m = 500 kg

We have

a =  \frac{6000}{500}  =  \frac{60}{5}  = 12 \\

We have the final answer as

<h3>12 m/s²</h3>

Hope this helps you

8 0
2 years ago
Two 4.3546 cm x 4.3546 cm square aluminum electrodes, spaced 0.6408 mm apart are connected to a 73.68 V battery. What is the cap
Helen [10]

Answer:

Capacitance, C = 26.1 picofarad

Explanation:

It is given that,

Side of square, x = 4.3546 cm = 0.043546 m

Distance between electrodes, d = 0.6408 mm = 0.0006408 m

Voltage, V = 73.68 V

Capacitance of parallel plates is given by :

C=\dfrac{\epsilon_oA}{d}

C=\dfrac{8.85\times 10^{-12}\times (0.043546)^2}{0.0006408}

C=2.61\times 10^{-11}\ F

or

C = 26.1 picofarad

So, the capacitance of the capacitor is 26.1 picofarad. Hence, this is the required solution.

3 0
3 years ago
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