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gizmo_the_mogwai [7]
3 years ago
6

What is the specific fuel requirement for flight under VFR at night in an airplane?

Physics
1 answer:
MrMuchimi3 years ago
6 0

Answer:

A). Enough to fly to the first point of intended landing and to fly after that for 45 minutes at normal cruising speed

Explanation:

<u>Here are Fuel requirements for flight in VFR conditions</u>

No person may begin a flight in an airplane under VFR conditions unless  there is enough fuel to fly to the first point of intended landing and, assuming normal cruising speed -

  • During the day, to fly after that for at least 30 minutes; or
  • At night, to fly after that for at least 45 minutes.
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The rocket's acceleration has components \(a_{x}(t)= \alpha t^{2}\) and \(a_{y}(t)= \beta - \gamma t\), where \(\alpha = 2.50 {\
lbvjy [14]
 it is just a matter of integration and using initial conditions since in general dv/dt = a it implies v = integral a dt 
v(t)_x = integral a_{x}(t) dt = alpha t^3/3 + c the integration constant c can be found out since we know v(t)_x at t =0 is v_{0x} so substitute this in the equation to get v(t)_x = alpha t^3 / 3 + v_{0x} 
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3 years ago
You are on a 4 m high ladder and throw a ball upwards at 12 m/s. It lands on the ground below the ladder.
Gelneren [198K]

a) 2.75 s

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y= h+ut-\frac{1}{2}gt^2

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g = 9.8 m/s^2 is the acceleration of gravity (downward)

We can find the time t at which the ball reaches the ground by substituting y=0 into the equation:

0 = 4 + 12t - 4.9 t^2

This is a second-order equation. By solving it for t, we find:

t = -0.30 s

t = 2.75 s

The first solution is negative, so we discard it; the second solution, t = 2.75 s, is the one we are looking for.

b) -15.0 m/s (downward)

The final velocity of the ball can be calculated by using the equation:

v=u-gt

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3 years ago
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Explanation:

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Explanation:

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