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nydimaria [60]
2 years ago
10

What is the minimum runway length that will serve? hint: you can solve this problem using ratios without having any additional i

nformation. express your answer in miles.
Physics
1 answer:
ehidna [41]2 years ago
6 0

There are many factors that determine if an aircraft can operate from a given airport. Of course the availability of certain services, such as fuel, access to air stairs and maintenance are all necessary. But before considering anything else, one must determine if the plane can physically land at an airport, and equally as important, take off.

What is the minimum runway length that will serve?

Looking at aerial views of runways can lead some to the assumption that they are all uniform, big and appropriate for any plane to land. This couldn’t be further from the truth.

A given aircraft type has its own individual set of requirements in regards to these dimensions. The classic 150’ wide runway that can handle a wide-body plane for a large group charter flight isn’t a guarantee at every airport. Knowing the width of available runways is important for a variety of reasons including runway illusion and crosswind condition.

Runways also have different approach categories based on width, and have universal threshold markings that indicate the actual width.

To learn more about runway

brainly.com/question/11553726

#SPJ4

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Flapping flight is very energy intensive. A wind tunnel test
steposvetlana [31]

Answer:

The metabolic power for starting flight=134.8W/kg

Explanation:

We are given that

Mass of starling, m=89 g=89/1000=0.089 kg

1 kg=1000 g

Power, P=12 W

Speed, v=11 m/s

We have to find the metabolic power for starting flight.

We know that

Metabolic power for starting flight=\frac{P}{m}

Using the formula

Metabolic power for starting flight=\frac{12}{0.089}

Metabolic power for starting flight=134.8W/kg

Hence, the metabolic power for starting flight=134.8W/kg

4 0
3 years ago
A truck with 28-in.-diameter wheels is traveling at 50 mi/h. Find the angular speed of the wheels in rad/min, *hint convert mile
solong [7]

Answer:

Angular speed ω=3771.4 rad/min

Revolution=5921 rpm

Explanation:

Given data

d=28in\\r=d/2=28/2=14in\\v=50mi/hr

To find

Angular speed ω

Revolution per minute N

Solution

First we need to convert the speed of truck to inches per mile

as

1 mile=63360 inches

1 hour=60 minutes

so

v=(50*\frac{63360}{60} )\\v=52800in/min

Now to solve for angular speed ω by substituting the speed v and radius r in below equation

w=\frac{v}{r}\\ w=\frac{52800in/min}{14in}\\ w=3771.4rad/min

To solve for N(revolutions per minute) by substituting the angular speed ω in the following equation

N=\frac{w}{2\pi }\\ N=\frac{3771.4rad/min}{2\pi }\\ N=5921RPM  

3 0
3 years ago
A 20-newton weight is attached to a spring.
Makovka662 [10]

Answer:

40 N/m

Explanation:

The diagram attached is used to answer the question

We know from Hooke's law that extension is directly proportional to the applied force hence

F=kx where x is extension, F is applied force and k is the spring constant. Making k the subject of the formula then

k=\frac {F}{k}

From the attached diagram extension is given by subtracting unstretched spring from stretched spring hence extension, x=1-0.5=0.5m

Substituting 20 N for F and 0.5 m for x then

k=\frac {20}{0.5}=40 N/m

8 0
3 years ago
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