Answer:
Surface tension in water
Friction between tires and pavement
Dissolution of salt in water
Explanation:
Surface tension in water: It is due to the electrostatic force of attraction (cohesive force) between water molecules.
Friction between tires and pavement: It is due to the attractive force between tires and pavement.
Dissolution of salt in water: The ions of
and
separate due to the strong attraction of water molecules.
V = 8 * 10^2 km/h = 800km/h
S= 1,8* 10^3 km = 1800km
t = ?
v = S/t
t = S/v
t = 1800km/ 800km/h
t ≈ 2,25h (135min)
Gain in decibels is given by;
Gain db = 10*log (Po/Pi), where Po = Power output, Pin = Power input
Substituting;
Gain in db = 10 * log (50/5) = 10 db
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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KE = (1/2)·(mass)·(speed)²
It doesn't matter whether the object is in a car, on a boat,
falling, on a conveyor belt, or being carried by ants.
KE = (1/2)·(40 kg)·(2 m/s)²
KE = (20 kg)·(4 m²/s²)
KE = 80 kg-m²/s²
KE = 80 Joules