Example 3 :
1 answer:
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:
The additional charge that will flow from the battery onto the positive plate is 0.924 nC
Explanation:
Given;
Area of the capacitor, A = 15.0 cm² = 15 x 10⁻⁴ m²
Separation distance, d = 3.00 mm = 3 x 10⁻³ m²
voltage of the capacitor, V = 58.0 V
dielectric constant, k = 4.60
Initial Capacitance of the capacitor before the addition of dielectric material:
Initial charge across the parallel plates:
Q₁ = CV
Q = 4.425 x 10⁻¹² x 58 = 2.567 x 10⁻¹⁰ C
Capacitance of the capacitor after the addition of dielectric material:
Cequ. = Ck
Cequ. = 4.425 x 10⁻¹²F x 4.6 = 20.355 x 10⁻¹² F
Final charge across the parallel plates:
Q₂ = Cequ. x V
Q₂ = 20.355 x 10⁻¹² x 58 = 11.806 x 10⁻¹⁰ C
Additional charge = Q₂ - Q₁
= 11.806 x 10⁻¹⁰ C - 2.567 x 10⁻¹⁰ C
= 9.239 x 10⁻¹⁰ C
= 0.924 nC
Therefore, the additional charge that will flow from the battery onto the positive plate is 0.924 nC