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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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An object is traveling on a circle with a radius of 6 feet. If in 80 seconds a central angle of 9/4 radians is swept out, then f
trapecia [35]

Answer:

The angular speed of the object is 0.0281 rad/s

The linear speed of the object is 0.169 ft/s

Explanation:

Given;

radius of the circle, r = 6 ft

time of motion of the object around the circle, t = 80 s

central angle formed by the object during the motion, θ = 9/4 rad = 2.25 rad

The angular speed of the object is calculated as;

\omega = \frac{\theta }{t} = \frac{2.25 \ rad}{80 \ s} = 0.0281 \ rad/s

The linear speed of the object is calculated as;

v = ωr

v = 0.0281 rad/s   x    6ft

v = 0.169 ft/s

8 0
3 years ago
A bullet is fired through a wooden board with a thickness of 9.0 cm. The bullet hits the board perpendicular to it, and with a s
gtnhenbr [62]

Answer:

-387883.3 m/s²

0.000286168546055 seconds

Explanation:

t = Time taken

u = Initial velocity = 370 m/s

v = Final velocity = 259 m/s

s = Displacement = 9 cm

a = Acceleration

v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{259^2-370^2}{2\times 9\times 10^{-2}}\\\Rightarrow a=-387883.3\ m/s^2

The acceleration of the bullet is -387883.3 m/s²

v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{259-370}{-387883.3}\\\Rightarrow t=0.000286168546055\ s

The time taken is 0.000286168546055 seconds

4 0
4 years ago
Space debris left from old satellites and their launchers is becoming a hazard to other satellites. (a) Calculate the speed of a
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Answer:

Part a)

v = 7407.1 m/s

Part b)

v_{rel} = 1.05 \times 10^4 m/s

Explanation:

Part a)

As we know that orbital velocity at certain height from the surface of Earth is given as

v = \sqrt{\frac{GM}{R+h}}

here we know that

M = 5.98 \times 10^{24} kg

R = 6.37 \times 10^6 m

h = 900 km = 9.0 \times 10^5 m

now we have

v = \sqrt{\frac{(6.67 \times 10^{-11})(5.98 \times 10^{24})}{6.37 \times 10^6 + 9.0 \times 10^5}}

v = 7407.1 m/s

Part b)

When a loose rivet is moving in same orbit but at 90 degree with the previous orbit path then in that case the relative speed of the rivet with respect to the satellite is given as

v_{rel} = \sqrt{2} v

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3 years ago
Please Help Quick ASAp Hurry
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Answer:

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