In this case the rubber raft has horizontal and vertical motion.
Considering vertical motion first.
We have displacement , u = Initial velocity, t = time taken, a = acceleration.
In vertical motion
s = 1960 m, u = 0 m/s, a = 9.81
So raft will take 20 seconds to reach ground.
Now considering horizontal motion of raft
u = 109 m/s, t = 20 s, a = 0
So
So shipwreck was 2180 meter far away from the plane when the raft was dropped.
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
Answer:
Explanation:
V=IR
R=V/I
R=9/0.0175