The flow rate on an arterial is 1,800 veh/h, evenly distributed over two lanes. If the average speed in these lanes is 40 mi/h,
1 answer:
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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:
surface temperature = 128.74⁰c
Explanation:
Given data
diameter of cable = 5 mm = 0.005 m
length of cable = 4 m
T∞ ( surrounding temperature ) = 20⁰c
voltage drop across cable ( dv )= 60 V
current across cable = 1.5 A
attached to this answer is the comprehensive analysis and solution to the problem.
The assumption made is not a good one since the calculated Ts ( surface temperature ) is very much different from the assumed Ts
