Answer:
8.8 cm
31.422 cm/s
Explanation:
m = Mass of block = 0.6 kg
k = Spring constant = 15 N/m
x = Compression of spring
v = Velocity of block
A = Amplitude
As the energy of the system is conserved we have
Amplitude of the oscillations is 8.8 cm
At x = 0.7 A
Again, as the energy of the system is conserved we have
The block's speed is 31.422 cm/s
Answer:
hello your question has some missing values attached below is the complete question with the missing values
answer :
a) 0.083 secs
b) 0.33 secs
c) 3e^-4/3
Explanation:
Given that
g = 32 ft/s^2 , spring constant ( k ) = 2 Ib/ft
initial displacement = 1 ft above equilibrium
mass = weight / g = 4/32 = 1/8
damping force = instanteous velocity hence β = 1
a<u>)Calculate the time at which the mass passes through the equilibrium position.</u>
time mass passes through equilibrium = 1/12 seconds = 0.083
<u>b) Calculate the time at which the mass attains its extreme displacement </u>
time when mass attains extreme displacement = 1/3 seconds = 0.33 secs
<u>c) What is the position of the mass at this instant</u>
position = 3e^-4/3
attached below is the detailed solution to the given problem
Given
v = 343 m/s
ac = 5g
ac = 5*9.8 m/s^2
ac = 49 m/s^2
where,
v: velocity
ac = centripetal aceleration
Procedure
We call the acceleration of an object moving in uniform circular motion—resulting from a net external force—the centripetal acceleration ac; centripetal means “toward the center” or “center seeking”.
Formula
The minimum radius not to exceed the centripetal acceleration is 2401 m.