The angular speed is decreasing and direction of rotation clockwise of the rod immediately after time t.
<h3>
</h3><h3>What is angular speed ?</h3>
The rate of change of angular displacement is defined as angular speed. It is stated as follows:
ω = θ t
Where,
θ is the angle of rotation,
t is the time
ω is the angular velocity
The torque is found as;l
If the force is acting on the rod from the three point is the same, the value of the torque is depends upon the radius or the perpendicular distance.
The perpendicular distance of the right force is grater. Hence, the force acting on the right side is more, and the rod will rotate clockwise.
Both the forces are acting downwards. Thus, the resultant force is the less due to which the speed is increasing.
Hence, the angular speed is decreasing and direction of rotation clockwise of the rod immediately after time t.
To learn more about the angular speed, refer to the link;
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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
<span>
Of course. Wind is air in motion, and the gases in air are composed of
all the usual familiar stuff ... atoms, molecules, mass, etc. That's how
the wind moves things ... it has momentum and kinetic energy, which
get transferred to the things that move in the wind.</span>
Answer:
a) 4.49Hz
b) 0.536kg
c) 2.57s
Explanation:
This problem can be solved by using the equation for he position and velocity of an object in a mass-string system:
for some time t you have:
x=0.134m
v=-12.1m/s
a=-107m/s^2
If you divide the first equation and the third equation, you can calculate w:
with this value you can compute the frequency:
a)
b)
the mass of the block is given by the formula:
c) to find the amplitude of the motion you need to know the time t. This can computed by dividing the equation for v with the equation for x and taking the arctan:
Finally, the amplitude is:
