Answer:
a true b false thos is the solution
Answer:
There are 50 ASE certification tests, covering almost every imaginable aspect of the automotive repair and service industry.
Explanation:
yww <33
Answer:
a) V(t) = Ldi(t)/dt
b) If current is constant, V = 0
Explanation:
a) The voltage, V(t), across an inductor is proportional to the rate of change of the current flowing across it with time.
If V represents the Voltage across the inductor
and i(t) represents the current across the inductor in time, t.
V(t) ∝ di(t)/dt
Introducing a proportionality constant,L, which is the inductance of the inductor
The general equation describing the voltage across the inductor of inductance, L, as a function of time when a current flows through it is shown below.
V(t) = Ldi(t)/dt ..................................................(1)
b) If the current flowing through the inductor is constant i.e. does not vary with time
di(t)/dt = 0 and hence the general equation (1) above becomes
V(t) = 0
Answer:
526.5 KN
Explanation:
The total head loss in a pipe is a sum of pressure head, kinetic energy head and potential energy head.
But the pipe is assumed to be horizontal and the velocity through the pipe is constant, Hence the head loss is just pressure head.
h = (P₁/ρg) - (P₂/ρg) = (P₁ - P₂)/ρg
where ρ = density of the fluid and g = acceleration due to gravity
h = ΔP/ρg
ΔP = ρgh = 1000 × 9.8 × 7.6 = 74480 Pa
Drag force over the length of the pipe = Dynamic pressure drop over the length of the pipe × Area of the pipe that the fluid is in contact with
Dynamic pressure drop over the length of the pipe = ΔP = 74480 Pa
Area of the pipe that the fluid is in contact with = 2πrL = 2π × (0.075/2) × 30 = 7.069 m²
Drag Force = 74480 × 7.069 = 526468.1 N = 526.5 KN
Answer:
230.4W
Explanation:
Heat transfer by conduction consists of the transport of energy in the form of heat through solids, in this case a jacket.
the equation is as follows
Where
Q=heat
k=conductivity=0.04
A=Area=1.8m^2
T2=33C
T1=1C
L=thickness=1cm=0.01m
Q=230.4W
the skier loses heat at the rate of 230.4W
