The question is incomplete. The complete question is :
A viscoelastic polymer that can be assumed to obey the Boltzmann superposition principle is subjected to the following deformation cycle. At a time, t = 0, a tensile stress of 20 MPa is applied instantaneously and maintained for 100 s. The stress is then removed at a rate of 0.2 MPa s−1 until the polymer is unloaded. If the creep compliance of the material is given by:
J(t) = Jo (1 - exp (-t/to))
Where,
Jo= 3m^2/ GPA
to= 200s
Determine
a) the strain after 100's (before stress is reversed)
b) the residual strain when stress falls to zero.
Answer:
a)-60GPA
b) 0
Explanation:
Given t= 0,
σ = 20Mpa
Change in σ= 0.2Mpas^-1
For creep compliance material,
J(t) = Jo (1 - exp (-t/to))
J(t) = 3 (1 - exp (-0/100))= 3m^2/Gpa
a) t= 100s
E(t)= ΔσJ (t - Jo)
= 0.2 × 3 ( 100 - 200 )
= 0.6 (-100)
= - 60 GPA
Residual strain, σ= 0
E(t)= Jσ (Jo) ∫t (t - Jo) dt
3 × 0 × 200 ∫t (t - Jo) dt
E(t) = 0
Given parameters:
Mass of puck = 162g = 0.162kg (1000g = 1kg)
Force exerted on puck = 171N
Final velocity = 42.3m/s
Unknown
A. time of the acceleration
B. radius of the curve?
Solution:
A. time of the acceleration
the initial velocity of the puck = 0m/s
We know that;
Force = mass x acceleration
Acceleration =
Acceleration =
So force = mass x
Input the parameters and solve for time;
171 = 0.162 x
171 =
t = = 0.04s
The time of acceleration is 0.04s
B. radius of the curve;
to solve this, we apply the centripetal force formula;
F =
where;
F is the centripetal force
m is the mass
v is the velocity
r is the radius
Since the force exerted on the puck is 151;
input the parameters and solve for r²;
151 =
151r = 0.162 x 42.3²
r = 1.92m
The radius of the circular curve is 1.92m
D thymine
if this is high school than i am tottaly right
One nanometer is equal to 1e-7 centimeters