A current value of 0.000005 A is equal to 5 HA.
What is Current?Current is a measure of rate. In physics, there are various rate quantities. For instance, velocity, which measures how quickly an item changes position, is a rate quantity.
Velocity is defined mathematically as the position change to time ratio. The rate at which an object changes its velocity is called acceleration, which is a rate quantity.
Acceleration is defined mathematically as the ratio of velocity change to time. Power, which measures the rate at which work is done on an object, is a rate quantity.
A current value of 0.000005 A is equal to 5 HA.
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There are gauges available for checking panel flushness. TRUE OR FALSE?
Answer: true
Explanation:
because flushness means your dead so you are dead now
A jet aircraft is in level flight at an altitude of 30,000 ft with an airspeed of 500 ft/s. The aircraft has a gross weight of 19,815 lb, a wingspan of 53.3 ft, and an average chord length of 6 ft. The Oswald efficiency factor is 0.81 and the zero-lift drag coefficient is equal to 0.02. The jet has two turbofan engines, each producing a maximum thrust of 3,650 lb at sea level.
Required:
a. Create a plot of the drag polar for this aircraft for CL from 0 to 5. Plot CL on the vertical axis, CD on the horizontal axis, and do not include negative CL values.
b. What is the total drag coefficient at the flight condition described above?
c. What is the required thrust for level flight at this altitude in lb?
d. If the pilot runs the engines at maximum thrust, what is the instantaneous rate of climb at this altitude and velocity?
Answer:
a) attached below
b) 0.0337
c) 2730.206 Ib
d) 2320.338 ft/min
Explanation:
a) Plot of the drag polar for this aircraft
first we will calculate :
Wing area (s) = Wing span (b) * Average chord length(c)
= 53.3 * 6 = 319.8 ft^2
Aspect ratio = b^2 / s = 8.883
K = 1 / [tex]\pi[/tex]eAR = 1 /
Drag polar ( Cd ) = 0.02 + 0.044 C^2L
attached below is a plot of the drag polar
Attached below is the detailed solution of the remaining part of the question
7.13 An intersection approach has a saturation flow rate of 1500 veh/h, and vehicles arrive at the approach at the rate of 800 veh/h. The approach is controlled by a pretimed signal with a cycle length of 60 seconds and D/D/1 queuing holds. Local standards dictate that signals should be set such that all approach queues dissipate 10 seconds before the end of the effective green portion of the cycle. Assuming that approach capacity exceeds arrivals, determine the maximum length
Answer:
23.34 seconds
Explanation:
Flow rate = 1500
Arrival = 800 vehicle per hour
Cycle c = 60 seconds
Dissipation time = 10 seconds
Arrival time = 800/3600 = 0.2222
Rate of departure = 1500/3600 = 0.4167
Traffic density p = 0.2222/0.4167 = 0.5332
Real time = r
r + to + 10 = c
to = c-r-10 ----1
t0 = p*r/1-p ----2
Equate both 1 and 2
C-r-10 = p*r/1-p
60-r-10 = 0.5332r/1-0.5332
50-r = 0.5332r/0.4668
50-r = 1.1422r
50 = 1.1422r + r
50 = 2.1422r
r = 50/2.1422
r = 23.34 seconds
Q.13 In order to produce maximum starting torque in a split-phase motor, how many degrees out of phase should the start- and run-winding currents be with each other?
Select one:
A. 180°
B. 0°
C. 120°
D. 90°
Answer:
D. 90 degrees.
Explanation:
Torque is a rotational force which moves an object in other direction. There should be 90 degrees out of phase to start, run winding currents with each other. Torque is produced by the rotational motion of an object. The angle of the object must be 90 degrees set in order to create torque.