Answer: I hope this helps you
Explanation:
Answer:
Explanation:
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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.
Explanation:
Elongation of the wire is:
ΔL = F L₀ / (E A)
where F is the force,
L₀ is the initial length,
E is Young's modulus,
and A is the cross sectional area.
ΔL = T (0.5 m) / ((2.0×10¹¹ Pa) (0.02 cm²) (1 m / 100 cm)²)
ΔL = T (1.25×10⁻⁶ m/N)
T = (80,000 N/m) ΔL
Draw a free body diagram of the mass at the bottom of the circle. There are two forces: tension force T pulling up and weight force mg pulling down.
Sum of forces in the centripetal direction:
∑F = ma
T − mg = mv²/r
T − mg = mω²r
T − (15 kg) (9.8 m/s²) = (15 kg) (2 rev/s × 2π rad/rev)² (0.5 m + ΔL)
T − 147 N = (2368.7 N/m) (0.5 m + ΔL)
Substitute:
(80,000 N/m) ΔL − 147 N = (2368.7 N/m) (0.5 m + ΔL)
(80,000 N/m) ΔL − 147 N = 1184.35 N + (2368.7 N/m) ΔL
(797631.3 N/m) ΔL = 1331.35 N
ΔL = 0.00167 m
ΔL = 1.67 mm
