Answer:
<em>1.228 x </em>
<em> mm </em>
<em></em>
Explanation:
diameter of aluminium bar D = 40 mm
diameter of hole d = 30 mm
compressive Load F = 180 kN = 180 x 
N
modulus of elasticity E = 85 GN/m^2 = 85 x 
Pa
length of bar L = 600 mm
length of hole = 100 mm
true length of bar = 600 - 100 = 500 mm
area of the bar A = 
= 
= 1256.8 mm^2
area of hole a = 
= 
= 549.85 mm^2
Total contraction of the bar = 
total contraction = 
==> 
= <em>1.228 x </em><em> mm </em>
Answer:
R = 2216m and The normal force of the seat on the pilot is 5008N
Explanation:
See attachment below please.
Answer:
Be pushed away from each other.
Explanation:
Good morning.
We have that:
, since we have rest in the inicial time.
The acceleration can be found with Newton's Law:
Now we put the acceleratin in the velocity equation:
We want the force, so, let's isolate
F:
