The answer to that question is c. tamod
A. NaOH
B. PbNO3
C. FeSO4
D. P2O3
E. CSe2
F. HC2H3O2
G. HCIO
H. H2SO4
3.
a)
r = distance of each mass in each hand from center = 0.6 m
m = mass of each mass in each hand = 2 kg
v = linear speed = 1.1 m/s
L = combined angular momentum of the masses = ?
Combined angular momentum of the masses is given as
L = 2 m v r
L = 2 (2) (1.1) (0.6)
L = 2.64 kg m²/s
b)
v' = linear speed when she pulls her arms = ?
r' = distance of each mass from center after she pulls her arms = 0.15 m
Using conservation of momentum , angular momentum remains same, hence
L = 2 m v' r'
2.64 = 2 (2) (0.15) v'
v' = 4.4 m/s
Answer: T = 472.71 N
Explanation: The wire vibrates thus making sound waves in the tube.
The frequency of sound wave on the string equals frequency of sound wave in the tube.
L= Length of wire = 26cm = 0.26m
u=linear density of wire = 20g/m = 0.02kg/m
Length of open close tube = 86cm = 0.86m
Sound waves in the tube are generated at the second vibrational mode, hence the relationship between the length of air and and wavelength is given as
L = 3λ/4
0.86 = 3λ/4
3λ = 4 * 0.86
3λ = 3.44
λ = 3.44/3 = 1.15m.
Speed of sound in the tube = 340 m/s
Hence to get frequency of sound, we use the formulae below.
v = fλ
340 = f * 1.15
f = 340/ 1.15
f = 295.65Hz.
f = 295.65 = frequency of sound wave in pipe = frequency of sound wave in string.
The string vibrated at it fundamental frequency hence the relationship the length of string and wavelength is given as
L = λ/2
0.26 = λ/2
λ = 0.52m
The speed of sound in string is given as v = fλ
Where λ = 0.52m f = 295.65 Hz
v = 295.65 * 0.52
v = 153.738 m/s.
The velocity of sound in the string is related to tension, linear density and tension is given below as
v = √(T/u)
153.738 = √T/ 0.02
By squaring both sides
153.738² = T / 0.02
T = 153.738² * 0.02
T = 23,635.372 * 0.02
T= 472.71 N
