Answer:
Output signal shape: square, from 0.1 to 230 MHz. Output power: -10 dBm (at a load of 50 Ohms).
Explanation:
If both the ram air input and drain hole of the pitot system become blocked, the indicated airspeed will: a) increase during a climb.
<h3>What is a
ram air input?</h3>
A ram air input can be defined as an air intake system which is designed and developed to use the dynamic air pressure that is created due to vehicular motion, or ram pressure, in order to increase the static air pressure within the intake manifold of an internal combustion engine of an automobile.
This ultimately implies that, a ram air input allows a greater mass-flow of air through the engine of an automobile, thereby, increasing the engine's power.
In conclusion, indicated airspeed will increase during a climb when both the ram air input and drain hole of the pitot system become blocked.
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Complete Question:
If both the ram air input and drain hole of the pitot system become blocked, the indicated airspeed will
a) increase during a climb
b) decrease during a climb
c) remain constant regardless of altitude change
The answer is B because it could be feasible but it’s not a need it and you got a time frame but it’s not a requirement and it doesn’t have to be unique.
Answer:
Time taken for the capacitor to charge to 0.75 of its maximum capacity = 2 × (Time take for the capacitor to charge to half of its capacity)
Explanation:
The charging of a capacitor/the build up of its voltage follows an exponential progression and is given by
V(t) = V₀ [1 - e⁻ᵏᵗ]
where k = (1/time constant)
when V(t) = V₀/2
(1/2) = 1 - e⁻ᵏᵗ
e⁻ᵏᵗ = 0.5
In e⁻ᵏᵗ = In 0.5 = - 0.693
-kt = - 0.693
kt = 0.693
t = (0.693/k)
Recall that k = (1/time constant)
Time to charge to half of max voltage = T(1/2)
T(1/2) = 0.693 (Time constant)
when V(t) = 0.75
0.75 = 1 - e⁻ᵏᵗ
e⁻ᵏᵗ = 0.25
In e⁻ᵏᵗ = In 0.25 = -1.386
-kt = - 1.386
kt = 1.386
t = 1.386(time constant) = 2 × 0.693(time constant)
Recall, T(1/2) = 0.693 (Time constant)
t = 2 × T(1/2)
Hope this Helps!!!
