Answer:
leak in cylinder head, piston holes, bad valves.
Explanation:
TAS is an important input which is required from ADC to allow the INS calculate W/V.
<h3>What is
INS?</h3>
INS is abbreviation for Inertial Navigation System and it can be defined as a navigation device that makes use of motion sensors, a computer, and rotation sensors, so as to continuously calculate by dead reckoning the velocity, position, and orientation of a moving object.
In the Aviation and Engineering filed, True Airspeed (TAS) is an important input which is required from ADC to allow the Inertial Navigation System (INS) calculate W/V.
Read more on Inertial Navigation System here: brainly.com/question/26052911
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Answer:
Permanent felony conviction on your criminal record.
Financial restitution if DUI involves an accident, property damage or personal injury
Up to 50 hours community service or fee of $10 per hour of service.
Minimum fine of upto $1000 or $2000, that is if your blood alcohol level was 0.15% or more than this, or you were driving with a minor in the vehicle.
Imprisonment of upto 9 months, if the blood alcohol level was over 0.15% or idf the DUI resulted in ana accident or crash, jail time is increased to one year.
Mandatory imprisonment for not less than 10 days if your second DUI conviction occurs within 5 years of the date of the prior DUI conviction.
180 days to one year probation.
Maximum of 5 years driver's license revocation.
Up to 30 days impoundment of vehicle.
Ignition interlock device.
Criminal record stating adjudication of guilt.
Explanation:
Answer:
(N-1) × (L/2R) = (N-1)/2
Explanation:
let L is length of packet
R is rate
N is number of packets
then
first packet arrived with 0 delay
Second packet arrived at = L/R
Third packet arrived at = 2L/R
Nth packet arrived at = (n-1)L/R
Total queuing delay = L/R + 2L/R + ... + (n - 1)L/R = L(n - 1)/2R
Now
L / R = (1000) / (10^6 ) s = 1 ms
L/2R = 0.5 ms
average queuing delay for N packets = (N-1) * (L/2R) = (N-1)/2
the average queuing delay of a packet = 0 ( put N=1)
Answer:
248.756 mV
49.7265 µA
Explanation:
The Thevenin equivalent source at one terminal of the bridge is ...
voltage: (100 V)(1000/(1000 +1000) = 50 V
impedance: 1000 || 1000 = (1000)(1000)/(1000 +1000) = 500 Ω
The Thevenin equivalent source at the other terminal of the bridge is ...
voltage = (100 V)(1010/(1000 +1010) = 100(101/201) ≈ 50 50/201 V
impedance: 1000 || 1010 = (1000)(1010)/(1000 +1010) = 502 98/201 Ω
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The open-circuit voltage is the difference between these terminal voltages:
(50 50/201) -(50) = 50/201 V ≈ 0.248756 V . . . . open-circuit voltage
__
The current that would flow is given by the open-circuit voltage divided by the sum of the source resistance and the load resistance:
(50/201 V)/(500 +502 98/201 +4000) = 1/20110 A ≈ 49.7265 µA
