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m_a_m_a [10]
3 years ago
7

A Cessna aircraft has a liftoff speed of 120 km/h What minimum constant acceleration does the aircraft require to be airborne af

ter a takeoff run of 240 m
Physics
1 answer:
Lady_Fox [76]3 years ago
4 0

Answer:

<h3>2.3125m/s²</h3>

Explanation:

Using the equation of motion v² = u²+2aS

v is the final velocity = 120km/hr

120km/hr = 120 * 1000/1 * 3600 = 33.3m/s

u is the initial velocity = 0m/s

a is the acceleration

S is the distance covered = 240m

On substituting the given parameters

33.3² = 0²+2a(240)

33.3² = 480a

1110 = 480a

a = 1110/480

a = 2.3125m/s²

Hence the minimum constant acceleration that the aircraft require to be airborne after a takeoff run of 240 m is 2.3125m/s²

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Harman [31]

It is dangerous to stand close to railroad tracks when a rapidly moving commuter train passes. Explain why atmospheric pressure would push you toward the moving train. Water pressure inside a hose nozzle can be less than atmospheric pressure due to the Bernoulli effect.

Pls brainliest!? :)

6 0
3 years ago
¿Por qué en algunas partes el agua tiene mayor presión?
Phantasy [73]
Cuando la presión aumenta en el agua, disminuye el punto de fusión del hielo. Ósea esta a temperatura muy alta.


Espero que esto te ayude y no me haya confundido
7 0
2 years ago
A horse has an acceleration of 2 m/s2. If it starts from rest, how fast is it going after 1.7 seconds?
fgiga [73]

       (2  m/s²)  ·  (1.7  s)  =  3.4 m/s
6 0
3 years ago
Read 2 more answers
A steel ball bearing with a radius of 1.5 cm forms an image of an object that has been placed 1.1 cm away from the bearing’s sur
Nonamiya [84]

Answer:

Check the explanation

Explanation:

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

The image is virtual

The image is upright

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

Kindly check the diagram in the attached image below.

5 0
3 years ago
As she climbs a hill a cyclist slows down from 25 mi/hr to 6mi/hr in 10 seconds what is her acceleration
vesna_86 [32]

|Acceleration| = (change in speed) / (time for the change).

Change in speed = (6 mi/hr - 25 mi/hr) = -19 mi/hr
Time for the change = 10 sec

|Acceleration| = (-19 mi/hr) / (10 sec) = -1.9 mile per hour per second

Admittedly, that's a rather weird unit.
Other units, perhaps more comfortable ones, are:

                 -6,840 mi/hr²

                   -2.79 feet/sec²

8 0
3 years ago
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