Answer:
3.14 × 10⁻⁴ m³ /s
Explanation:
The flow rate (Q) of a fluid is passing through different cross-sections remains of pipe always remains the same.
Q = Area x velocity
Given:
Diameters of 3 sections of the pipe are given as
d1 = 1.0 cm, d2 = 2.0 cm and d3 = 0.5 cm.
Speed in the first segment of the pipe is
v1 = 4 m/s.
From the equation of continuity the flow rate through different cross-sections remains the same.
Flow rate = Q = A1 v1 = A2 v2 = A3 v3.
Q = A1v1
=π/4 d²1 v1 = π/4 * 0.01² ×4.0 m³/s = 3.14 × 10⁻⁴ m³ /s
The voltage across an inductor ' L ' is
V = L · dI/dt .
I(t) = I(max) sin(ωt)
dI/dt = I(max) ω cos(ωt)
V = L · ω · I(max) cos(ωt)
L = 1.34 x 10⁻² H
ω = 2π · 60 = 377 /sec
I(max) = 4.80 A
V = L · ω · I(max) cos(ωt)
V = (1.34 x 10⁻² H) · (377 / sec) · (4.8 A) · cos(377 t)
<em>V = 24.25 cos(377 t)</em>
V is an AC voltage with peak value of 24.25 volts and frequency = 60 Hz.
300 000 0 squared = 2 x 9.8 distance
KINEMATICS
Uniform or constant motion in a straight line (rectilinear). Speed or velocity constant and/or acceleration constant. If motion is up and down and/or has an up and down component then acceleration omn earth will be g. g is about 10m/s/s.
speed = distance/time
velocity = displacement/time
s=distance ... u=initial speed ... v = final speed ... a = acceleration ... t = time
v=u+at
v^2=u^2+2as
s=ut+1/2at^2
