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3 years ago

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How do the structures of the respiratory system work together to supply my body with oxygen and get rid of carbon dioxide? do no

t look it up

1 answer:

AlexFokin [52]3 years ago

5 0

A

Explanation:

I hate tiny links OMLJDJDJDJSJDJJDJDJ

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Piano tuners tune pianos by listening to the beats between the harmonics of two different strings. When properly tuned, the note

Sophie [7]

(a) 2 Hz

The frequency of the nth-harmonic is given by

$f_n = n f_1$

where

$f_1$ is the fundamental frequency

Therefore, the frequency of the third harmonic of the A ( $f_1 = 440 Hz$ ) is

$f_3 = 3 \cdot f_1 = 3 \cdot 440 Hz =1320 Hz$

while the frequency of the second harmonic of the E ( $f_1 = 659 Hz$ ) is

$f_2 = 2 \cdot f_1 = 2 \cdot 659 Hz =1318 Hz$

So the frequency difference is

$\Delta f = 1320 Hz - 1318 Hz = 2 Hz$

(b) 2 Hz

The beat frequency between two harmonics of different frequencies f, f' is given by

$f_B = |f'-f|$

In this case, when the strings are properly tuned, we have

- Frequency of the 3rd harmonic of A-note: 1320 Hz

- Frequency of the 2nd harmonic of E-note: 1318 Hz

So, the beat frequency should be (if the strings are properly tuned)

$f_B = |1320 Hz - 1318 Hz|=2 Hz$

(c) 1324 Hz

The fundamental frequency on a string is proportional to the square root of the tension in the string:

$f_1 \propto \sqrt{T}$

this means that by tightening the string (increasing the tension), will increase the fundamental frequency also*, and therefore will increase also the frequency of the 2nd harmonic.

In this situation, the beat frequency is 4 Hz (four beats per second):

$f_B = 4 Hz$

And since the beat frequency is equal to the absolute value of the difference between the 3rd harmonic of the A-note and the 2nd harmonic of the E-note,

$f_B = |f_3-f_2|$

and $f_3 = 1320 Hz$ , we have two possible solutions for $f_2$ :

$f_2 = f_3 - f_B = 1320 Hz - 4 Hz = 1316 Hz\\f_2 = f_3 + f_B = 1320 Hz + 4 Hz = 1324 Hz$

However, we said that increasing the tension will increase also the frequency of the harmonics (*), therefore the correct frequency in this case will be

1324 Hz

8 0

2 years ago

A block of wood is floating in water; it is depressed slightly and then released to oscillate up and down. Assume that the top a

Marysya12 [62]

Explanation:

Equilibrium position in y direction:

W = Fb (Weight of the block is equal to buoyant force)

m*g = V*p*g

V under water = A*h

hence,

m = A*h*p

Using Newton 2nd Law

$-m*\frac{d^2y}{dt^2} = Fb - W\\\\-m*\frac{d^2y}{dt^2} = p*g*(h+y)*A - A*h*p*g\\\\-A*h*p*\frac{d^2y}{dt^2} = y *p*A*g\\\\\frac{d^2y}{dt^2} + \frac{g}{h} * y =0$

Hence, T time period

T = 2*pi*sqrt ( h / g )

4 0

3 years ago

If a plane is moving at a constant velocity what is happening to the acceleration?

quester [9]

The plane is not accelerating.

Hope this helps!

4 0

2 years ago

Read 2 more answers

A 6 N and a 10 N force act on an object. The moment arm of the 6 N force is 0.2 m. If the 10 N force produces five times the tor

Levart [38]

Answer:

The moment arm is 0.6 m

Explanation:

Given that,

First force $F_{1}=6\ N$

Second force $F_{2}=10\ N$

Distance r = 0.2 m

We need to calculate the moment arm

Using formula of torque

$\tau=Force\times lever\ arm$

So, Here,

$\tau_{2}=5 \tau_{1}$

We know that,

The torque is the product of the force and distance.

Put the value of torque in the equation

$F_{2}\times d_{2}=5\times F_{1}\times r_{1}$

$r_{2}=\dfrac{5\times F_{1}\times r_{1}}{F_{2}}$

Where, $F_{1}$ =First force

$F_{1}$ =First force

$F_{2}$ =Second force

$r_{1}$ = distance

Put the value into the formula

$r_{2}=\dfrac{5\times6\times0.2}{10}$

$r_{2}=0.6\ m$

Hence, The moment arm is 0.6 m

6 0

3 years ago

Bella makes the 2.5m distance to her food bowl in 9.1 seconds. What is her average velocity?

e-lub [12.9K]

Distance=2.5m
Time=9.1s

Average Velocity=Total Distance/Total Time

$\\ \sf\longmapsto \dfrac{2.5}{9.1}$

$\\ \sf\longmapsto 0.3m/s$

7 0

2 years ago