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3 years ago

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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How is the surface area of an average person is 2m^2 in the chapter pressure

11Alexandr11 [23.1K]

Answer:

We have to show surface area $=2m^2$ ,with few conditions that is by considering Force $=200000\ N$ and Pressure $=100000\ Pa$ to be respectively.

Explanation:

The atmospheric pressure is $=10^{5}\ Pa$ on Earth's surface.

The magnitude of the force exerted on a person by the atmosphere is $=2\times 10^{5}\N (or)\ 200kN\ (or)\ 20\ ton$ .

Now to calculate surface area we can find it from $pressure=\frac{force}{area}$ and re-arranging it to.

$area=\frac{force}{pressure}$

So plugging the values,

Surface area $=\frac{20000}{10000}=2\ m^{2}$

Hence from the above calculations we can say that surface area is $2m^2$ .

So the surface area of an average person can be said to have $2m^2$ , using the concept of pressure and force.

3 0

3 years ago

Read 2 more answers

An engine absorbs 1749 J from a hot reservoir and expels 539 J to a cold reservoir in each cycle. a. What is the engine’s effici

Masja [62]

Answer:

a)η = 69.18 %

b)W= 1210 J

c)P=3967.21 W

Explanation:

Given that

Q₁ = 1749 J

Q₂ = 539 J

From first law of thermodynamics

Q₁ = Q₂ +W

W=Work out put

Q₂=Heat rejected to the cold reservoir

Q₁ =heat absorb by hot reservoir

W= Q₁- Q₂

W= 1210 J

The efficiency given as

$\eta=\dfrac{W}{Q_1}$

$\eta=\dfrac{1210}{1749}$

$\eta=0.6918$

η = 69.18 %

We know that rate of work done is known as power

$P=\dfrac{W}{t}$

$P=\dfrac{1210}{0.305}\ W$

P=3967.21 W

8 0

3 years ago

A cannonball is launched from the ground at an angle of 30 degrees above the horizontal and a speed of 30 m/s. Ideally (no air r

marusya05 [52]

As we know that here no air resistance while ball is moving in air

So here we will say that

initial total energy = final total energy

$KE_i + U_i = KE_f + U_f$

here we know that

$Ui = U_f = 0$ (as it will be on ground at initial and final position)

so we will say

$KE_i = KE_f$

since mass is always conserved

so we will say that final speed of the ball must be equal to the initial speed of the ball

so we have

$v_f = v_i = 30 m/s$

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3 years ago

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A 35-mm single lens reflex (SLR) digital camera is using a lens of focal length 35.0 mm to photograph a person who is 1.80 m tal

olganol [36]

Answer:

a) 35.44 mm

b) 17.67 mm

Explanation:

u = Object distance = 3.6 m

v = Image distance

f = Focal length = 35 mm

$h_u$ = Object height = 1.8 m

a) Lens Equation

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{35}-\frac{1}{3600}\\\Rightarrow \frac{1}{v}=\frac{713}{25200} \\\Rightarrow v=\frac{25200}{713}=35.34\ mm$