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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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A particle oscillates in simple harmonic motion, with amplitude A and period T. The particle starts from position x = A. What is
frozen [14]

Answer:

i guess it is D because,

y = A sin(wt)

w = \frac{2\pi }{T} and t = 3T/ 2

now, Y = A sin ( \frac{2\pi }{T} X \frac{3T}{2})

so, Y = 0radian

Explanation:

8 0
3 years ago
According to the graph, during which time interval are the particles in the air slowing down?
kenny6666 [7]
Answer b like the last time should be it
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1 year ago
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A solid cylinder with a mass of 2.46 kg and a radius of 0.049 m starts from rest at a height of 4.80 m and rolls down a 24.3 ◦ s
irina1246 [14]

Answer:

v_{f}\approx 2.097\,\frac{m}{s}

Explanation:

Let assume that the solid cylinder rolls down a frictionless incline. The translational speed can be found by using the Principle of Energy Conservation and the Work-Energy Theorem:

m_{cyl}\cdot g\cdot y_{o} = \frac{1}{2}\cdot m_{cyl} \left( 1+ \frac{1}{R}\right)\cdot v_{f}^{2}

g\cdot y_{o} = \frac{1}{2}\cdot\left( 1+ \frac{1}{R}\right)\cdot v_{f}^{2}

The translational speed is:

v_{f} = \sqrt{\frac{2\cdot g\cdot y_{o}}{\left(1 + \frac{1}{R}  \right)}}

v_{f} = \sqrt{\frac{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (4.80\,m)}{\left(1 + \frac{1}{0.049\,m}  \right)} }

v_{f}\approx 2.097\,\frac{m}{s}

7 0
3 years ago
A 12-V battery is connected across a 100-Ω resistor. How many electrons flow through the wire in 1.0 min?
Vanyuwa [196]

Answer:

The quantity of charge or electron flowing the wire in the given time is 4.5 x 10¹⁹ electrons.

Explanation:

Given;

emf of the battery, V = 12 V

resistance of the resistor, R = 100-Ω

time of current flow, t = 1 min

charge of 1 electron = 1.602 x 10¹⁹ C

The current through this circuit is given by;

I = V / R

I = (12) / (100)

I = 0.12 A

The quantity of charge or electron flowing the wire in the given time is calculated as;

Q =It

where;

I is the current flowing through the wire

t is the time of current flow = 1 x 60s = 60 s

Q = 0.12 x 60

Q = 7.2 C

1.602 x 10⁻¹⁹ C --------------- 1 electron

7.2 C -----------------------------? electron

= \frac{7.2 }{1.602*10^{-19}} \\\\= 4.5*10^{19} \ electrons

Therefore, the quantity of charge or electron flowing the wire in the given time is 4.5 x 10¹⁹ electrons.

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2 years ago
Which of the following is a common property of Noble Gases?
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<span>Have a full outer shell of electrons.</span>
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