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babunello [35]
3 years ago
10

During a cross-country flight you picked up rime icing which you estimate is 1/2" thick on the leading edge of the wings. You ar

e now below the clouds at 2000 feet AGL and are approaching your destination airport under VFR. Visibility under the clouds is more than 10 miles, winds at the destination airport are 8 knots right down the runway, and the surface temperature is 3 degrees Celsius. You decide to ____.
Physics
1 answer:
igor_vitrenko [27]3 years ago
6 0

Answer:

Use a faster than normal approach and landing speed.

Explanation

For pilots, it is one of the critical moments of the flight that concentrates 12% of fatal accidents. The main difficulty lies in reaching enough speed to take flight within the space of the runway. At present, it ceased to be a challenge for the aircraft, since the engine power improved, so the takeoff ceased to be the most dangerous moment of the flight.

One of the risks that aircraft face today is that some of the engines fail while the plane accelerates. In that case, the pilot must decide in an instant whether it is better to take flight and solve the problem in the air or if it is preferable not to take off.

Although for many staying on the ground might seem the most sensible option, it is not as simple as it seems: to suddenly decelerate an aircraft, with the weight it has and the speed it reaches can cause accidents. However, today a special cement was designed that runs around the runways of the airports, which when coming into contact with the wheels of the aircraft the ground breaks and helps to slow down.

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brainly a child on a swing set swings back and forth with a period of 3.3 s and an amplitude of 25°. what is the maximum speed o
beks73 [17]

Maximum speed of the child as she swings is  2.23 m/s.

<h3>Step by Step Calculation:</h3>

T=3.3 s is the oscillation's time period.

The swing's greatest angle is 25° (max).

The swing's bottom will have the following kinetic energy:

k=12mv2...........(1)

The mass in this situation is m, and the speed is v.

The potential energy change is expressed as,

∆U=mgL1-cosθmax...............(2)

Here, L is a string's length and g is the acceleration caused by gravity. L is given as,

L=gT24π2

Combine equation (1) with (2)

12mv2=mgL1-cos, maxv=2g, maxv=gT24, maxv=g2T22, maxv=9.8 m/s

22.33 m/s, 22.31 s22.31 cos25°

Therefore, the child's top speed is 2.23 m/s.

<h3>What is Oscillation ?</h3>
  • The process of any quantity or measure fluctuating repeatedly about its equilibrium value in time is known as oscillation.
  • A periodic change in a substance's value between two values or around its central value is another way to define oscillation.

To learn more about Oscillation refer to:

brainly.com/question/28312746

#SPJ1

6 0
1 year ago
A worker leaves home at 9:00 AM, travels 40 km to the office, and then returns home at 5:00 PM. What is the magnitude of the wor
saveliy_v [14]
The answer is 0 km 

brainlist please!
7 0
3 years ago
Read 2 more answers
A vehicle reaches a speed of 7.5 m/s over 15 seconds. What is its acceleration if it
denpristay [2]

acceleration = \frac{velocity}{time}  = \frac{7.5}{15}  = 0.5ms^-^2

So the answer is option b.

8 0
3 years ago
What is a Vector and a scalar?
vfiekz [6]
Scalars are quantities that are fully described by a magnitude (or numerical value) alone.
Vectors are quantities that are fully described by both a magnitude and a direction.
4 0
3 years ago
Q11) If you were standing at the top of a building and you dropped a rock.
Dafna1 [17]

Answer:

Part A

The distance travel by the rock is approximately 132.496 m

Part B

The speed when the rock hits the ground is approximately 50.96 m/s

Explanation:

Part A

The question is focused on the kinetics equation of a free falling object

The given parameter is the time it takes the rock to hit the ground, t = 5.2 s

For an object in free fall, we have;

h = 1/2·g·t²

Where;

h = The height from which the object is dropped

g = The acceleration due to gravity ≈ 9.8 m/s²

t = The time taken to travel the distance, h = 5.2 s

∴ h = 1/2 × 9.8 m/s² × (5.2 s)² ≈ 132.496 m

The distance travel by the rock, h ≈ 132.496 m

Part B

The speed, 'v', when the rock hits the ground, is given by the following kinematic equation,

v = g·t

∴ v = 9.8 m/s² × 5.2 s = 50.96 m/s

The speed when the rock hits the ground, v ≈ 50.96 m/s.

8 0
2 years ago
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