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Katarina [22]
3 years ago
11

The lowest-pitch tone to resonate in a pipe of length L that is closed at one end and open at the other end is 200 Hz. Which one

of the following frequencies will NOT resonate in the same pipe
a. 1800 Hz
b. 1000 Hz
c. 1400 Hz
d. 600 Hz
e. 400 Hz
Physics
1 answer:
ozzi3 years ago
7 0

Answer:

e. 400 Hz

Explanation:

In closed organ pipe,  only odd harmonics of fundamental note is possible .

The fundamental frequency is 200 Hz . Then other overtones will be having following frequencies .

200 x 3 , 200 x 5 , 200 x 7 , 200 x 9 etc

600 Hz , 1000 Hz , 1400 Hz  , 1800 Hz .

Frequency not possible is 400 Hz .

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We want to find how much charge is on the electrons in a nickel coin. Follow this method. A nickel coin has a mass of about 4.2
monitta

Answer:

The number of atoms is N = 4.37*10^{22} \ atoms

Explanation:

From the question we are told that

                 The mass of coin  is m_n = 4.2g

                   Number of atom in one mole = n =6.02*10^{23} \ atoms

                   Molar mass of nickel M = 57.8g

Now the relation to obtain the number of atom in  the  nickel coin is

                        N = \frac{Mass \ of Nickel\ coin}{Molar \ mass\ of nickel }  * No\ of\atoms \ in \ \ one\  mole\ of\ nickel

                           = \frac{4.2}{57.8}* 6.02*10^{23}

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8 0
3 years ago
A frictionless piston-cylinder device contains 10 kg of superheated vapor at 550 kPa and 340oC. Steam is then cooled at constant
Alla [95]

Answer:

a) the work (W) done during the process is -2043.25 kJ

b) the work (W) done during the process is -2418.96 kJ

Explanation:

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mass of water vapor m = 10 kg

initial pressure P₁ = 550 kPa

Initial temperature T₁ = 340 °C

steam cooled at constant pressure until 60 percent of it, by mass, condenses; x = 100% - 60% = 40% = 0.4

from superheated steam table

specific volume v₁ = 0.5092 m³/kg

so the properties of steam at p₂ = 550 kPa, and dryness fraction

x = 0.4

specific volume v₂ = v_f + xv_{fg

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( The negative, indicates work is done on the system )

b)

What would the work done be if steam were cooled at constant pressure until 80 percent of it, by mass, condenses

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Now, work done during the process will be;

W = mP₁( v₂ - v₁ )

W = 10 × 550( 0.06939 - 0.5092 )

W = 5500 × -0.43981

W = -2418.96 kJ

Therefore, the work (W) done during the process is -2418.96 kJ

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