The capacitance of a Pt-n-type GaAs Schottky diode is given by 1 (C(μF))2 = 1.0 × 105 − 2.0 × 105 V The diode area is 0.1 cm2. C
alculate the built-in voltage Vbi, the barrier height, and the doping concentration
1 answer:
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Answer:
Δt =
Explanation:
Hi there!
Using the equation of speed for the whole trip, we can obtain the time each one needed to cover the distance D.
The speed (v) is calculated by dividing the traveled distance (d) over the time needed to cover that distance (t):
v = d/t
Rick traveled half of the distance at Vr and the other half at Vw. Then, when v = Vr, the distance traveled was D/2 and the time is unknown, Δt1:
Vr = D/ (2 · Δt1)
For the other half of the trip the expression of velocity will be:
Vw = D/(2 · Δt2)
The total time traveled is the sum of both Δt:
Δt(total) = Δt1 + Δt2
Then, solving the first equation for Δt1:
Vr = D/ (2 · Δt1)
Δt1 = D/(2 · Vr)
In the same way for the second equation:
Δt2 = D/(2 · Vw)
Δt + Δt2 = D/(2 · Vr) + D/(2 · Vw)
Δt(total) = D/2 · (1/Vr + 1/Vw)
The time needed by Rick to complete the trip was:
Δt(total) = D/2 · (1/Vr + 1/Vw)
Now let´s calculate the time it took Tim to do the trip:
Tim walks half of the time, then his speed could be expressed as follows:
Vw = 2d1/Δt Where d1 is the traveled distance.
Solving for d1:
Vw · Δt/2 = d1
He then ran half of the time:
Vr = 2d2/Δt
Solving for d2:
Vr · Δt/2 = d2
Since d1 + d2 = D, then:
Vw · Δt/2 + Vr · Δt/2 = D
Solving for Δt:
Δt (Vw/2 + Vr/2) = D
Δt = D / (Vw/2 + Vr/2)
Δt = D/ ((Vw + Vr)/2)
Δt = 2D / (Vw + Vr)
The time needed by Tim to complete the trip was:
Δt = 2D / (Vw + Vr)
Let´s find the diference between the time done by Tim and the one done by Rick:
Δt(tim) - Δt(rick)
2D / (Vw + Vr) - (D/2 · (1/Vr + 1/Vw))
= Δt
Let´s check the result. If Vr = Vw:
Δt = 2D/2Vr - D/2Vr - D/2Vr
Δt = D/Vr - D/Vr = 0
This makes sense because if both move with the same velocity all the time both will do the trip in the same time.
Answer:
Explanation:
As we projected the balloon at speed vo at an angle of 29 degree
so the two component of velocity is given as
now we know that in x direction we have
in y direction we have
now we have