The friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 × 10^8 respectively. Also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 × 10^8 respectively.
<h3>How to determine the friction factor</h3>
Using the formula
μ = viscosity = 0. 06 Pas
d = diameter = 120mm = 0. 12m
V = velocity = 1m/s and 3m/s
ρ = density = 0.9
a. Velocity = 1m/s
friction factor = 0. 52 ×
friction factor = 0. 52 ×
friction factor = 0. 52 × 0. 55
friction factor
b. When V = 3mls
Friction factor = 0. 52 ×
Friction factor = 0. 52 ×
Friction factor = 0. 52 × 0. 185
Friction factor
Loss When V = 1m/s
Head loss/ length = friction factor × 1/ 2g × velocity^2/ diameter
Head loss = 0. 289 × × ×
Head loss = 1. 80 × 10^8
Head loss When V = 3m/s
Head loss = × × ×
Head loss = 5. 3× 10^8
Thus, the friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 ×10^8 respectively also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 ×10^8 respectively.
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Answer:
7. (D) uniformly accelerated vertical motion
8. (A) zero
9. (A) zero
10. (C) parabolic
According to the conservation of energy
- Potential energy at any given instance is equal to the Kinetic energy as energy can neither be created nor be destroyed
Mass is 2kg=m
#A
h=5m
PE
Now
- KE=98J
- 1/2mv²=98J
- 1/2×2v²=98J
- v²=98J
- v=√98
- v=9.4m/s
#B
h=3.2m
PE:-
Now
- KE=62.7J
- 1/2mv²=62.7
- v²=62.7
- v=√62.7
- v=7.9m/s
#C
h=2m
PE
Now
Answer: 13 m/s
Explanation: The two vectors form a 5-12-13 right triangle. the magnitude of their resultant is the hypotenuse, which is 13 m/s.