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Answer:
pressure of the water = 3.3 × pa
Explanation:
given data
velocity v1 = 1.5 m/s
pressure P = 400,000 Pa
inside radius r1 = 1.00 cm
pipe radius r2 = 0.5 cm
h1 = 0 (datum at inlet)
h2 = 5.0 m (datum at inlet)
density of water ρ = 1000 kg/m³
to find out
pressure of the water
solution
we consider here flow speed in bathroom that is = v2 and Pressure in bathroom is = P2
here we will use both continuity and Bernoulli equations
because here we have more than one unknown so that
v1 × A1 = v2 × A2 × P1 + ρ g h1 + (0.5)ρ v1² = P2 + ρ g h2 + (0.5) ρ v2²
now we use here first continuity equation for get v2
v2 =
v2 =
v2 = 6 m/s
and now we use here bernoulli eqution for find here p2 that is
P2 = P1 - 0.5× ρ ×(v2² - v1²) - ρ g (h2- h1 )
P2 = 400000 - 0.5× 1000 ×(6² - 1.5²) - 1000 × 9.81 × (5-0 )
P2 = 3.3 × pa
Answer:
Maximum force will be equal to 720 N
Explanation:
We have given that spring constant
Maximum stretch of the spring x = 6 cm = 0.06 m
We have to find the maximum force on the spring
We know that spring force is given by
So the maximum force which is necessary to relaxed the spring will be eqaul to 720 N