A hose pipe has a cross sectional area of 10cm^2 and Water of density 1000kg / m^3 is flowing out of ito at Speed tomis at a rate of o.5kg/s , find the force required to hold th pipe steady?
Answers
Volume of water =1500 litres =1500∗1000 cm
3
=1500000 cm
3
Cross-section area 2 cm
2
Time =5 min=5×60=300 sec
Volume is given in 5 min through 2cm
2
cross section area hose pipe.
So,
Speed of water = Volume of water/cross-section area*time
S=
2×300
1500000
=2500 cm/s=25m/s
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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