Answer:
Explanation:
mass attached m = .14 kg
force constant k = 5N / m
displacement
= amplitude of oscillation
A = .03 m
A ) period of motion = 
= 2 x 3.14 
T = 1.05 s
B ) maximum speed of block = angular velocity x amplitude
= (2π /T) x A
= (2 x 3.14 x .03) / 1.05
= .1794 m / s
17.94 cm /s
C )
maximum acceleration = angular velocity² x amplitude
= (2π /T)² x A
= (2π /1.05)² x .03
= 1.073 m / s²
D )
position
S = A cos ωt , ω is angular velocity
S = .03 cos(2πt /T)
= .03 cos 5.98 t
v =ω A sin(2πt /T)
= 5.98 x .03 sin5.98t
= .1794 sin5.98t
acceleration = ω²A sin5.98t
= 1.073 sin5.98t
Answer:

Explanation:
Given that,
The half-life of Barium-139 is 
A sample contains
nuclei.
We need to find the decay constant for this decay. The formula for half life is given by :

Put all the values,

So, the decay constant is
.