A hose pipe of cross section area 10cm^2 and a water of density 1000kg/ m^3 is flowing out of it at a speed 10m/s .Find the force required to hold the pipe steady .
Answers
Area = 10cm root 2
Water of density = 1000kg/ m root 3
speed = 10m/s
The force required to hold the pipe steady
1m = 100 s
10m = 1000 s
1000 m/ s / 10m/s
10 ( Ans). ( this is a force)
A = F/ A
= 10 / 10 cm root cube is answer is cm3.
✌✌
The force required to hold the pipe steady = 100 N
Step-by-step explanation:
Given:
Cross section area of a hose pipe = 10 cm²
Density of water = 1000 kg/m³
Speed of water coming out of hose pipe = `10 m/s
To find out:
The force required to hold the pipe steady
Solution:
A = 10 cm²
ρ = 1000 kg/m³
v = 10 m/s
Volume of water coming out per second
Q = vA = 10 × 10 × 10⁻⁴ m³/s
= 10⁻² m³/s
Mass of water coming out per second = Q×ρ
= 10⁻² × 1000 kg/s
= 10 kg/s
Therefore, change in momentum per second
= Q×ρ×v
= 10 × 10
= 100 kg-m/s/s
Thus, the force required to hold the pipe steady
F = Change in momentum
= Q×ρ×v
= 100 N
Hope this answer is helpful.
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