Ask question

Ask question

All categories

2 years ago

12

If you were to build a pipe organ with open-tube pipes spanning the range from 25 hz to 19 khz. The speed of sound in air is 343

m/s

1 answer:

Dennis_Churaev [7]2 years ago

3 0

The length of the shortest pipe which is l2 will be 0.018m

<h3>What is Length?</h3>

This is defined as the measurement or extent of a substance from end to end. The formula is shown below:

L = v/f where v is speed , f is frequency

f1 = 25hz , f2 = 19khz

The length for both sides is seen below:

l1 = 343/25 = 13.72m

l2 = 343/19000 = 0.018m.

The length of the shortest pipe is therefore 0.018m.

Read more about Length here brainly.com/question/24259483

You might be interested in

Explica de que tipo es cada oración según la actitud del hablante

Volgvan

Answer:

Según la actitud del hablante las oraciones se clasifican en enunciativas, interrogativas, etc. ... adverbios o expresiones que complementan a toda la oración (COr): ojalá, quizá.

Explanation:

7 0

2 years ago

Which properties of plastics allow them to be solutions to many complex problems in the world? Check all that apply. A.chemicall

grigory [225]

the answers are A, B, C, E HOPED THIS HELPED

3 0

3 years ago

Read 2 more answers

An inductor has inductance of 0.260 H and carries a current that is decreasing at a uniform rate of 18.0 mA/s.

nignag [31]

Answer:

The self-induced emf in this inductor is 4.68 mV.

Explanation:

The emf in the inductor is given by:

$\epsilon = -L\frac{dI}{dt}$

Where:

dI/dt: is the decreasing current's rate change = -18.0 mA/s (the minus sign is because the current is decreasing)

L: is the inductance = 0.260 H

So, the emf is:

$\epsilon = -L\frac{dI}{dt} = -0.260 H*(-18.0 \cdot 10^{-3} A/s) = 4.68 \cdot 10^{-3} V$

Therefore, the self-induced emf in this inductor is 4.68 mV.

I hope it helps you!

6 0

3 years ago

A 60-kg cheetah reaches a speed of 30 m/s as it chases its prey.What is the kinetic energy of the cheetah?

Rama09 [41]

The answer would be 27,000 Joules because (1/2) m v^2 =30*900 which equals 27,000 J

4 0

3 years ago

Read 2 more answers

A piston-cylinder device contains Helium gas initially at 150 kPa, 20 o C, and 0.5m 3 . The helium is now compressed in a polytr

Molodets [167]

Answer:

Explanation:

Given

$P_1=150 kPa$

$T_1=20^{\circ}C$

$V_1=0.5 m^3$

$T_2=140^{\circ}C$

$P_2=400 kPa$

R for Helium $R=2.076$

$c_v=3.115 kJ/kg-K$

mass of gas $m=\frac{P_1V_1}{RT_1}$

$m=\frac{150\times 0.5}{2.076\times 293}$

$m=0.123 kg$

Similarly $V_2$ can be found

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

$V_2=0.264 m^3$

Work done $W=\int_{V_1}^{V_2}PdV$

$W=\frac{P_2V_2-P_1V_1}{n-1}$

$W=\frac{mR(T_2_T_1)}{n-1}$

Since it is a polytropic Process

therefore $PV^n=c$

$P_1V_1^n=P_2V_2^n$

$(\frac{V_1}{V_2})^n=\frac{P_2}{P_1}$

$(\frac{0.5}{0.264})^n=\frac{400}{150}$

$n=\frac{\ln 2.66}{\ln 1.893}$

$n=1.533$

$W=\frac{0.123\times 2.076(140-20)}{1.533-1}$

$W=57.48 kJ$

From Energy balance

$E_{in}-E_{out}=\Delta E_{system}$

Neglecting kinetic and Potential Energy change

$Q_{in}+W_{in}=change\ in\ Internal\ Energy$

Change in Internal Energy $\Delta U=u_2-u_1$

$\Delta U=mc_v(T_2-T_1)$

$\Delta U=0.123\times 3.115(140-20)$

$\Delta U=45.977 kJ$

$Q_{in}+57.48=45.977$

$Q_{in}=-11.50 kJ$

i.e. Heat is being removed

3 0

3 years ago