Answer:
Option (d) 2 min/veh
Explanation:
Data provided in the question:
Average time required = 60 seconds
Therefore,
The maximum capacity that can be accommodated on the system, μ = 60 veh/hr
Average Arrival rate, λ = 30 vehicles per hour
Now,
The average time spent by the vehicle is given as
⇒ 
thus,
on substituting the respective values, we get
Average time spent by the vehicle = 
or
Average time spent by the vehicle = 
or
Average time spent by the vehicle = 
or
Average time spent by the vehicle = 
hr/veh
or
Average time spent by the vehicle = 
min/veh
[ 1 hour = 60 minutes]
thus,
Average time spent by the vehicle = 2 min/veh
Hence,
Option (d) 2 min/veh
Answer:
is this a question for hoework
Answer:
5984.67N
Explanation:
A 14 inch diameter pipe is decreased in diameter by 2 inches through a contraction. The pressure entering the contraction is 28 psi and a pressure drop of 2 psi occurs through the contraction if the upstream velocity is 4.0 ft/sec. What is the magnitude of the resultant force (lbs) needed to hold the pipe in place?
from continuity equation
v1A1=v2A2
equation of continuity
v1=4ft /s=1.21m/s
d1=14 inch=.35m
d2=14-2=0.304m
A1=pi*d^2/4
0.096m^2
a2=0.0706m^2
from continuity once again
1.21*0.096=v2(0.07)
v2=1.65
force on the pipe
(p1A1- p2A2) + m(v2 – v1)
from bernoulli
p1 + ρv1^2/2 = p2 + ρv2^2/2
difference in pressure or pressure drop
p1-p2=2psi
13.789N/m^2=rho(1.65^2-1.21^2)/2
rho=21.91kg/m^3
since the pipe is cylindrical
pressure is egh
13.789=21.91*9.81*h
length of the pipe is
0.064m
AH=volume of the pipe(area *h)
the mass =rho*A*H
0.064*0.07*21.91
m=0.098kg
(193053*0.096- 179263.6* 0.07) + 0.098(1.65 – 1.21)
force =5984.67N
In metallurgy, non-ferrous metals are metals or alloys that do not contain iron in appreciable amounts. Generally more costly than ferrous metals, non-ferrous metals are used because of desirable properties such as low weight, higher conductivity, non-magnetic property or resistance to corrosion
