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Zarrin [17]
1 year ago
12

determine the approximate time, compass heading, distance, and fuel consumed during climb cruising altitude 6,500

Physics
1 answer:
Maslowich1 year ago
8 0

The approximate time, compass heading, distance, and fuel consumed during the climb cruising altitude of 6,500 are <u>17 minutes, 224°, 36 NM, and 3.7 gallons.</u>

<h3>What is compass heading?</h3>
  • The compass heading that the bow or nose of the ship is pointing is known as the heading of a vessel or an aircraft in navigation.
  • Keep in mind that the compass heading may not always correspond to the vehicle's actual route or track, which is the direction it actually moves.
  • The motion of the underlying medium, such as the air or water, or other factors like skidding or slipping, is the cause of any discrepancy between the compass heading and course.
  • The wind triangle can be used to calculate the drift, which is the difference. There have been described at least seven different methods of measuring a vehicle's heading.

To learn more about compass heading with the given link

brainly.com/question/12704364

#SPJ4

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Answer:

m = 2.23 \times 10^{-32} kg

Explanation:

Given data:

PERIOD OF MOTION IS T = 25.5 days

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we know that

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Gravitational forceF= \frac{Gm m}{d^2}

we know that

v = \frac{2\pi R}{T}

solving for

R = \frac{vT}{2\pi}

F = \frac{Gm^2}{4(\frac{vT}{2\pi})^2}

F = G\times \frac{\pi m}{(vT)^2}

a = \frac{v^2}{\frac{vT}{2\pi}}

a = \frac{2\pi v}{T}

we know that

f =ma

G\times \frac{\pi m}{(vT)^2} = a = \frac{2\pi m v}{T}

solving for m

m = \frac{2Tv^3}{\pi G}

m = \frac{2\times 25.5 \times 86400 \times 220000^3\ m/s}{\pi \times 6.67\times 10^{-11}}

m = 2.23 \times 10^{-32} kg

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A locomotive accelerates a 25-car train along a level track. Every car has a mass of 7.7 ✕ 104 kg and is subject to a friction f
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To solve the problem it is necessary to apply the concepts related to Force of Friction and Tension between the two bodies.

In this way,

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Therefore the friction force at 29Km / h would be,

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