Answer:
BC and DE
Explanation:
In the given figure, the velocity time graph is shown. We know that the area under v-t curve gives the displacement of the particle.
Area under AB,
Area under BC,
Area under CD,
Area under DE,
Area under EF,
So, form above calculations it is clear that, during BC and DE undergo equal displacement. Hence, the correct option is (c) "BC and DE = 4 meters".
Answer : The value of the constant for a second order reaction is,
Explanation :
The expression used for second order kinetics is:
where,
k = rate constant = ?
t = time = 17s
= final concentration = 0.0981 M
= initial concentration = 0.657 M
Now put all the given values in the above expression, we get:
Therefore, the value of the constant for a second order reaction is,
Answer:
a) the maximum shear stress τ the bar is 16T
/πd³
b) the angle of twist between the ends of the bar is 16tL² / πGd⁴
Explanation:
Given the data in the question, as illustrated in the image below;
d is the diameter of the prismatic bar of length AB
t is the intensity of distributed torque
(a) Determine the maximum shear stress tmax in the bar
Maximum Applied torque T_max = tL
we know that;
shear stress τ = 16T/πd³
where d is the diameter
so
τ = 16T
/πd³
Therefore, the maximum shear stress τ the bar is 16T
/πd³
(b) Determine the angle of twist between the ends of the bar.
let theta () be the angle of twist
polar moment of inertia = πd⁴/32
now from the second image;
lets length dx which is at distance of "x" from "B"
Torque distance x
T(x) = tx
Elemental angle twist = d = T(x)dx / G
so
d = tx.dx / G(πd⁴/32)
d = 32tx.dx / πGd⁴
so total angle of twist will be;
=
=
32tx.dx / πGd⁴
= 32t / πGd⁴
= 32t / πGd⁴ [ L²/2]
= 16tL² / πGd⁴
Therefore, the angle of twist between the ends of the bar is 16tL² / πGd⁴
