a 6-kg and a 4-kg ball are acted on by forces of equal size. if the large ball accelerates at 2 m/s2 what acceleration will the
2 answers:
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Answer:
L=0.654 m
Explanation:
<u>Concepts and Principles </u>
1- The speed of sound in air is expressed as a function of the temperature of air as follows:
v=(331 m/s)√(1+T_C/273°C) (1)
where 331 m/s is the speed of sound in air at temperature 0°C and Tc is the temperature of air in Celsius.
<u>Standing Wave Patterns in Pipes: </u>
A pipe open at both ends can have standing wave patterns with resonant frequencies:
f=v/λ=nv/2L n=1,2,3.........
where v is the speed of sound in air.
<u>Given Data </u>
f_1 (fundamental frequency of the flute) = 262 Hz
T (temperature of the air) = 20°C
The flute is open at both ends.
<u>Required Data </u>
We are asked to determine the length of the tube.
<u>Solution</u><u> </u>
The speed of sound in air at temperature T = 20°C is found from Equation (1):
v=(331 m/s)√(1+T_C/273°C)
=342.91 m/s
The fundamental frequency of the flute is found by substituting n = 1 into Equation (2):
f=v/2L
Solve for L:
L=v/2f_1
L=0.654 m
Answer:
1200 Sm^2mol^-1
Explanation:
Given data :
conductivity of water ( kwater ) = 76 mS m^-1 = 0.076 Sm^-1
conductivity of kcl (aq)( Kkcl ) = 1.1639 Sm^-1
Kkcl = 1.1639 - 0.076 = 1.0879 Sm^-1
Resistance = 33.21 Ω
where conductivity can be expressed as =
hence cell constant = conductivity * Resistance
= 1.0879 * 33.21 = 36.13m^-1
conductivity of CH3COOH ( kCH3COOH ) = 36.13 / 300
= 0.120 Sm^-1
<u>Determine the molar conductivity of acetic acid</u>
= ( kCH3COOH * 1000 ) / C
C = 0.1 mol dm
= (0.120 * 1000) / 0.1 = 1200 Sm^2mol^-1