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The 2-kg spool s fits loosely on the inclined rod forwhich the coefficient of static friction is s= 0.2. if thespool is located
0.25 m from a, determine the minimumand maximum constant speed the spool can have so that it does not slipalongthe rod.
1 answer:
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Answer:
lowest frequency = 535.93 Hz
distance between adjacent anti nodes is 4.25 cm
Explanation:
given data
length L = 32 cm = 0.32 m
to find out
frequency and distance between adjacent anti nodes
solution
we consider here speed of sound through air at room temperature 20 degree is approximately v = 343 m/s
so
lowest frequency will be = ..............1
put here value in equation 1
lowest frequency will be =
lowest frequency = 535.93 Hz
and
we have given highest frequency f = 4000Hz
so
wavelength = ..............2
put here value
wavelength =
wavelength = 0.08575 m
so distance = ..............3
distance =
distance = 0.0425 m
so distance between adjacent anti nodes is 4.25 cm
Answer:
0.1 m
Explanation:
F = Force exerted on spring = 3 N
k = Spring constant = 60 N/m
x = Displacement of the block
As the energy of the system is conserved we have
The position of the block is 0.1 from the initial position.