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
Answer: 4.7m/s²
Explanation:
According to newton's first law,
Force = mass × acceleration
Since we are given more the one force, we will take the resultant of the two vectors.
Mass = 2.0kg
F1+F2 = (3i-8j)+(5i+3j)
Adding component wise, we have;
F1+F2 = 3i+5i-8j+3j
F1+F2 = 8i-5j
Resultant of the sum of the forces will be;
R² = (8i)²+(-5j)²
Since i.i = j.j = 1
R² = 8²+5²
R² = 64+25
R² = 89
R = √89
R = 9.4N
Our resultant force = 9.4N
Substituting in the formula
F = ma
9.4 = 2a
a = 9.4/2
a = 4.7m/s²
Therefore, magnitude of the acceleration of the particle is 4.7m/s²
Answer:
16.8 lb is the force on the brake pad of one wheel.
Explanation:
Force applied on the piston = 
Area of the piston = 
Force applied on the brakes = 
Area of the brakes = 
Applying Pascal's law: 'For an incompressible fluid pressure at one surface is equal to the pressure at other surface'.
16.8 lb is the force on the brake pad of one wheel.
<h3>The metre is the length of the path travelled by light in vacuum during a time interval of 1299 792 458 of a second.</h3>
