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IrinaVladis [17]

3 years ago

7

An organ pipe closed at one end is 0.76 m long. (A) Determine its fundamental frequency. (B) What are the 3rd and 5th harmonics?

Note: Use 344 m/s as the speed of sound.

1 answer:

DanielleElmas [232]3 years ago

8 0

Answer:

A) 113.16 Hz

B) 339.47 Hz

565.79 Hz

Explanation:

Only odd harmonics are present in closed pipes that was why they asked for 3rd and 5th harmonics.

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3 years ago

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A solvent passes through a chromatography column in 3.0 minutes, but the solute requires 9.0 minutes.

Sergio [31]

Answer:

(a). 2

(b). 1/3

(c). 11.11

Explanation:

(a). k= (t₍s₎-t₍o₎)/t₍o₎...............(1)

where k= retention factor,

t₍o₎=solvent time, t₍s₎= solute time.

Given t₍s₎=9.0 Minutes, t₍o₎=3.0 minutes.

∴ k= (9-3)/3

k= 2.

(b). the fraction of time the solute spend in the mobile phase in the column is the ratio of the solvent time to the solute time. = t₍o₎/t₍s₎..........(2)

= 3/9

=1/3.

(c). K=k(Vm/Vs)................(3)

where K= partition coefficient, k= retention factor, Vm=volume of mobile phase, Vs= volume of stationary phase.

∴K = k(Vm/Vs)

k=2, and Vs=0.18Vm.

∴K = 2(Vm/0.18vm)

⇒K = 2/0.18

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4 0

3 years ago

41. City A lies 30 km directly south of city B. A bus, beginning at city A travels 50 km at 37° north of east to reach city C. H

Nana76 [90]

Answer:

B) 40 km West

Explanation:

Let position of B is origin so with respect of City B the position of A is given as

$r_A = 30 km$ south of city B

now Bus starts from City A and travels a distance of 50 km at 37 degree North of East to reach city C

so it is

$r_{CA} = 50(cos37\hat i + sin37 \hat j)$

$r_{CA} = 40\hat i + 30\hat j$

now the position of C is given with respect to A

in order to find the position of C with respect to B we can say

$r_C = r_{CA} + r_A$

$r_C = 40\hat i + 30\hat j + (-30 \hat j)$

$r_C = 40 \hat i$

so in order to reach at city B from city C bus has to travel 40 km towards West

7 0

3 years ago

Be sure to answer all parts. Without stratospheric ozone (O3), harmful solar radiation would cause gene alterations. Ozone forms

Ratling [72]

Answer:

for C-Cl bond $\lambda_1=3.66\times 10^{-9}\ m$

for O-O bond $\lambda_2=2.43\times 10^{-9}\ m$

Explanation:

We have the bond enegy for C-Cl, $BE_1=327\ kJ.mol^{-1}= 327000\times 1.66\times 10^{-22}\ J\ per\ molecule$

bond enegy for C-Cl, $BE_2=494\ kJ.mol^{-1}= 494000\times 1.66\times 10^{-22}\ J\ per\ molecule$

<u>Now as we know the energy of electromagnetic waves is given by:</u>

$E=h.\nu$

here E = BE

<u>Then for C-Cl:</u>

$327000\times 1.66\times 10^{-22}=6.63\times 10^{-34}\times \nu_1$

$\nu_1=8.18733\times 10^{16}\ Hz$

Now wavelength:

$\lambda_1=\frac{c}{\nu_1}$

$\lambda_1=\frac{3\times 10^8}{8.18733\times 10^{16}}$

$\lambda_1=3.66\times 10^{-9}\ m$

<u>For O2:</u>

$494000\times 1.66\times 10^{-22}=6.63\times 10^{-34}\times \nu_2$

$\nu_2=1.236863\times 10^{17}\ Hz$

Now wavelength:

$\lambda_2=\frac{c}{\nu_2}$

$\lambda_2=\frac{3\times 10^8}{1.236863\times 10^{17}}$

$\lambda_2=2.43\times 10^{-9}\ m$

8 0

3 years ago

At a certain location, wind is blowing steadily at 10 m/s. Determine the mechanical energy of air per unit mass and the power ge

Effectus [21]

Answer:

$X=3976.078202kW \approx 3976kW$

Explanation:

From the question we are told that

Wind speed $V_w=10m/s$

Turbine blade diameter $D=90m$

Air density $\tau=1.25kg/m^3$

Mechanical energy $K.E =0.05kJ/kg$

Generally the power generation potential of the wind turbine X is mathematically given as

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$m'=1.25 *\pi*90^2/4*10$

$m'=79521.56404kg/s \approx 79521.5kg/s$

Therefore

$X=m'*K.E$

$X=79521.56404*0.05$

$X=3976.078202kW \approx 3976kW$

$X=4.0MW$

4 0

3 years ago