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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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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
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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

6 0
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

The CCD sensor is 35.34 mm from the lens

b) Magnification

m=-\frac{v}{u}\\\Rightarrow m=-\frac{35.34}{3600}

m=\frac{h_v}{h_u}\\\Rightarrow -\frac{35.34}{3600}=\frac{h_v}{1800}\\\Rightarrow h_v=-\frac{35.34}{3600}\times 1800=-17.67\ mm

The person appears 17.67 mm tall on the sensor

7 0
3 years ago
when 10 similar coins are dropped into a graduated cylinder from 75 ml to 100ml. what is average volume of each coin?​
Alex787 [66]

Explanation:

Volume of 10 coins = 100ml - 75ml = 25ml

Volume of 1 coin = 25ml / 10 = 2.5ml

The average volume of each coin is 2.5ml.

8 0
2 years ago
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