What is the average distance (in terms of R) between the mobile on the fringe of the serving cell and the second and third tier
1 answer:
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Answer:
Explanation:
= Area of section 1 =
= Velocity of water at section 1 = 100 ft/min
= Specific volume at section 1 =
= Density of fluid =
= Area of section 2 =
Mass flow rate is given by
The mass flow rate through the pipe is
As the mass flowing through the pipe is conserved we know that the mass flow rate at section 2 will be the same as section 1
The speed at section 2 is .
Answer:
S = 0.5 km
velocity of motorist = 42.857 km/h
Explanation:
given data
speed = 70 km/h
accelerates uniformly = 90 km/h
time = 8 s
overtakes motorist = 42 s
solution
we know initial velocity u1 of police = 0
final velocity u2 = 90 km/h = 25 mps
we apply here equation of motion
u2 = u1 + at
so acceleration a will be
a =
a = 3.125 m/s²
so
distance will be
S1 = 0.5 × a × t²
S1 = 100 m = 0.1 km
and
S2 = u2 × t
S2 = 25 × 16
S2 = 400 m = 0.4 km
so total distance travel by police
S = S1 + S2
S = 0.1 + 0.4
S = 0.5 km
and
when motorist travel with uniform velocity
than total time = 42 s
so velocity of motorist will be
velocity of motorist =
velocity of motorist =
velocity of motorist = 42.857 km/h