If a 10-volt battery is placed across a 20-ohm resistance, the current will be ______ amperes. a-200 b-0.5 c-2
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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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1) The total mechanical energy of the rock is:
where U is the gravitational potential energy and K the kinetic energy.
Initially, the kinetic energy is zero (because the rock starts from rest, so its speed is zero), and the total mechanical energy of the rock is just gravitational potential energy. This is equal to
where
is the mass, 
is the gravitational acceleration and 
is the height.
Putting the numbers in, we find the potential energy
2) Just before hitting the ground, the potential energy U is zero (because now h=0), and all the potential energy of the rock converted into kinetic energy, which is equal to:
where v is the speed of the rock just before hitting the ground. Since the mechanical energy of the rock must be conserved, then the kinetic energy K before hitting the ground must be equal to the initial potential energy U of the rock:
3) For the work-energy theorem, the work W done by the gravitational force on the rock is equal to the variation of kinetic energy of the rock, which is:
where U is the gravitational potential energy and K the kinetic energy.
Initially, the kinetic energy is zero (because the rock starts from rest, so its speed is zero), and the total mechanical energy of the rock is just gravitational potential energy. This is equal to
where
Putting the numbers in, we find the potential energy
2) Just before hitting the ground, the potential energy U is zero (because now h=0), and all the potential energy of the rock converted into kinetic energy, which is equal to:
where v is the speed of the rock just before hitting the ground. Since the mechanical energy of the rock must be conserved, then the kinetic energy K before hitting the ground must be equal to the initial potential energy U of the rock:
3) For the work-energy theorem, the work W done by the gravitational force on the rock is equal to the variation of kinetic energy of the rock, which is: