Answer:
2.36 x 10^5 kg
Explanation:
radius of hose, r = 0.017 m
radius of underground pipe, R = 0.088 m
number of hoses, n = 3
velocity of water in underground pipe, V = 2.7 m/s
Let v is the velocity of water in each hose.
According to the equation of continuity
A x V = n x a x v
π R² x V = n x π x r² x v
0.088 x 0.088 x 2.7 = 3 x 0.017 x 0.017 x v
v = 24.12 m/s
(a) Amount of water poured onto a fire in one hour by all the three hoses
= n x a x v x density of water x time
= 3 x 3.14 x 0.017 x 0.017 x 24.12 x 1000 x 3600
= 2.36 x 10^5 kg
Thus, the amount of water poured onto the fire in one hour is 2.36 x 10^5 kg.
Given:
V1 = 4m3
T1 = 290k
P1 = 475 kpa = 475000 Pa
V2 = 6.5m3
T2 = 277K
Required:
P
Solution:
n = PV/RT
n = (475000 Pa)(4m3) / (8.314 Pa-m3/mol-K)(290k)
n = 788 moles
P = nRT/V
P = (788 moles)(8.314
Pa-m3/mol-K)(277K)/(6.5m3)
P = 279,204 Pa or 279 kPa
If the conductor isn't working it would be blocked
