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Akimi4 [234]
4 years ago
6

A clarinet sounds as a closed pipe. if a clarinet sounds a note with a pitch of 375 hz, what are the frequencies of the lowest t

hree harmonics produced by this instrument?
Physics
1 answer:
mote1985 [20]4 years ago
8 0
So mathematical harmonics are based around a divergent set of fractions. Sigma(1/n)
with the 1st harmonic being... well 1, or 1 full wavelength.The second harmonic is exactly 1/2 the wavelength of the 1st with the third being 1/3 the wavelength. As Wavelengths go down, frequencies go up in a perfect ratio.

Second Harmonic has double the Frequency of the 1st or base note. Third Harmonic is triple and so on.

So the Harmonic set of 375 is.
1. 375
2. 375×2=750
3. 375×3= 1125
.
.
.
etc (: I hope this helps.
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tigry1 [53]
5089 I believe is the correct answer
8 0
3 years ago
Suppose the mass of a fully loaded module in which astronauts take off from the Moon is 12,900 kg. The thrust of its engines is
Alexandra [31]

Answer:

Acceleration a=0.5 m/s²

Explanation:

Given data

Mass m=12,900 kg=1.29×10⁴kg

Thrust of engine F=28,000 N=2.8×10⁴N

gravitational acceleration g=1.67 m/s²

To find

Acceleration

Solution

As we know  that

W_{weigth}=mg\\ W=(1.29*10^{4}kg )(1.67m/s^{2} )\\W=21543N

The net force can be given as

F_{net}=F_{thrust}-W\\F_{net}=(2.8*10^{4}-21543)N\\   F_{net}=6457 N

From Newtons second law of motion we know that

F=ma\\a=F/m\\a=\frac{6457N}{12,900kg}\\ a=0.5m/s^{2}

6 0
3 years ago
A weightlifter curls a 30 kg bar, raising it each time a distance of 0.50 m .How many times must he repeat this exercise to burn
ludmilkaskok [199]

Answer:

N = 2141 times

Explanation:

Each time the work done to raise a given mass is

W = mgh

here we know that

m = 30 kg

h = 0.50 m

now we have

W = (30 kg)(9.81 m/s^2)(0.50)

W = 147.15 J

since it is just 25% of actual energy consumed as we know its efficiency is 25%

so we have total energy consumed in this way

E_{total} = \frac{147.15}{0.25}

E_{total} = 588.6 J

now if it took N number of times so burn the fat of a pizza then

N(588.6) = 1260 \times 10^3

N = 2141 times

4 0
3 years ago
A tube has a length of 0.025 m and a cross-sectional area of 6.5 x 10-4 m2. The tube is filled with a solution of sucrose in wat
Pepsi [2]

Answer:

The time required for sucrose transportation through the tube is 8.4319 sec.

Explanation:

Given:

L = 0.025 m

A = 6.5×10^-4 m^2

D = 5×10^-10 m^2/s

ΔC = 5.2 x 10^-3 kg/m^3

m = 5.7×10^-13 kg

Solution:

t = m×L / D×A×ΔC

t = (5.7×10^-13) × (0.025) / (5×10^-10)×(6.5×10^-4)×(5.2 x 10^-3)

t = 8.4319 sec.

5 0
3 years ago
*help FAST* !15 pts!
Ludmilka [50]
I believe a but none sound like they’d build a laser
5 0
3 years ago
Read 2 more answers
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