Answer:
a) the velocity of the implant immediately after impact is 20 m/s
b) the average resistance of the implant is 40000 N
Explanation:
a) The impulse momentum is:
mv1 + ∑Imp(1---->2) = mv2
According the exercise:
v1=0
∑Imp(1---->2) = F(t2-t1)
m=0.2 kg
Replacing:

if F=2 kN and t2-t1=2x10^-3 s. Replacing

b) Work and energy in the system is:
T2 - U(2----->3) = T3
where T2 and T3 are the kinetic energy and U(2----->3) is the work.

Replacing:

Answer: The net force in every bolt is 44.9 kip
Explanation:
Given that;
External load applied = 245 kip
number of bolts n = 10
External Load shared by each bolt (P_E) = 245/10 = 24.5 kip
spring constant of the bolt Kb = 0.4 Mlb/in
spring constant of members Kc = 1.6 Mlb/in
combined stiffness factor C = Kb / (kb+kc) = 0.4 / ( 0.4 + 1.6) = 0.4 / 2 = 0.2 Mlb/in
Initial pre load Pi = 40 kip
now for Bolts; both pre load Pi and external load P_E are tensile in nature, therefore we add both of them
External Load on each bolt P_Eb = C × PE = 0.2 × 24.5 = 4.9 kip
So Total net Force on each bolt Fb = P_Eb + Pi
Fb = 4.9 kip + 40 kip
Fb = 44.9 kip
Therefore the net force in every bolt is 44.9 kip
I think it’s is false I’m not that sure
Answer:
The speed of shaft is 1891.62 RPM.
Explanation:
given that
Amplitude A= 0.15 mm
Acceleration = 0.6 g
So
we can say that acceleration= 0.6 x 9.81

We know that

So now by putting the values



We know that
ω= 2πN/60
198.0=2πN/60
N=1891.62 RPM
So the speed of shaft is 1891.62 RPM.