Answer:
3.08m/s²
Explanation:
Given parameters:
Radius = 20m
Tangential velocity = 7.85m/s
Unknown:
Centripetal acceleration = ?
Solution:
Centripetal acceleration is the acceleration of a body along a circular path.
it is mathematically given as;
a =
v is the tangential velocity
r is the radius
a =
= 3.08m/s²
Answer:
Explanation:
Given
diameter 
density 
frequency 
Length of silk strand 
Velocity in the string is as follows

The expression for Fundamental Frequency




Squaring




Answer:
v=0.94 m/s
Explanation:
Given that
M= 5.67 kg
k= 150 N/m
m=1 kg
μ = 0.45
The maximum acceleration of upper block can be μ g.
a= μ g ( g = 10 m/s²)
The maximum acceleration of system will ω²X.
ω = natural frequency
X=maximum displacement
For top stop slipping
μ g =ω²X
We know for spring mass system natural frequency given as

By putting the values

ω = 4.47 rad/s
μ g =ω²X
By putting the values
0.45 x 10 = 4.47² X
X = 0.2 m
From energy conservation


150 x 0.2²=6.67 v²
v=0.94 m/s
This is the maximum speed of system.
Answer:
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