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Kitty [74]
3 years ago
6

An Airbus A350 is initially moving down the runway at 6.0 m/s preparing for takeoff. The pilot pulls on the throttle so that the

engines give the plane a constant acceleration of 1.8 m/s^2. The plane then travels a distance of 1500 m down the runway before lifting off. How long does it take from the application of the acceleration until the plane lifts off, becoming airborne?
Physics
1 answer:
alexdok [17]3 years ago
8 0

Answer:

t=67.7s

Explanation:

From this question we know that:

Vo = 6m/s

a = 1.8 m/s2

D = 1500m

And we also know that:

X=V_{o}*t + \frac{a*t^{2}}{2}   Replacing the known values:

1500=6t+0.9*t^{2}    Solving for t we get 2 possible answers:

t1 = -44.3s   and t2 = 67.7s    Since negative time represents an instant before the beginning of the movement, t1 is discarded. So, the final answer is:

t = 67.7s

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Elodia [21]

The block's potential energy at the top of the incline (at a height h from the horizontal surface) is equal to its kinetic energy at the bottom of the incline, so that

mgh = 1/2 mv²

where v is its speed at the bottom of the incline. It follows that

v = √(2gh)

If the incline is 20.4 m long, that means the block has a starting height of

sin(15°) = h/(20.4 m)   ⇒   h = (20.4 m) sin(15°) ≈ 5.2799 m

and so the block attains a speed of

v = √(2gh) ≈ 10.1728 m/s

The block then slides to a rest over a distance d. Kinetic friction exerts a magnitude F over this distance and performs an amount of work equal to Fd. By the work-energy theorem, this quantity is equal to the block's change in kinetic energy, so that

Fd = 0 - 1/2 mv²   ⇒   d = (-1293.58 J)/F

By Newton's second law, the net vertical force on the block as it slides is

∑ F [vertical] = n - mg = 0

where n is the magnitude of the normal force, so that

n = mg = (25 kg) g = 245 N

and thus the magnitude of friction is

F = -0.16 (245 N) = -39.2 N

(negative since it opposes the block's motion)

Then the block slides a distance of

d = (-1293.58 J) / (-39.2 N) ≈ 32.9994 m ≈ 33 m

5 0
2 years ago
A +2e charge is at the point (-1,0) mm in the x,y plane. A –e charge is at the point (0,1) mm. What is the electric field at the
Gennadij [26K]

Answer:

Let I and j be the unit vector along x and y axis respectively.

Electric field at origin is given by

E= kq1/r1^2 i + kq2/r2^2j

= 9*10^9*1.6*10^-19*/10^-6*(2i+ j)

= (2.88i + 1.44j)*10^-3 N/C

Force on charge= qE= 3*10^-19*1.6*(2.88i +1. 44 j) *10^-3

F= (1.382 i + 0.691 j) *10^-21

 

 Goodluck

Explanation:

4 0
1 year ago
Show the weight of the ladder and draw the missing Frictional force.
madreJ [45]
Since there is no friction between the ladder and the wall, there can be no vertical force component. That's the tricky part ;)

So to find the weight, divide the 100N <em>normal</em> force by earths gravitational acceleration, 9.8m/s^2

W =  \frac{N}{g}  =  \frac{100N}{9.8m/s^{2}} =  \frac{100}{9.8} = 10.2kg


Then;
Draw an arrow at the base of the ladder pointing towards the wall with a value of 30N, to show the frictional force.
5 0
2 years ago
If a particle's position is given by x=4-12t+3t^2, where t is in seconds and x is in meters, what is its velocity at t=1 second?
andreyandreev [35.5K]

Answer:

v = -6m/s

Explanation:

x=4-12t+3t^2

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For t = 1:

\frac{dx}{dt}=-6

3 0
3 years ago
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Tanzania [10]

Answer:

4.32

Explanation:

The centripetal acceleration of any object is given as

A(cr) = v²/r, where

A(c) = the centripetal acceleration

v = the linear acceleration

r = the given radius, 1.9 m

Since we are not given directly the centripetal acceleration, we'd be using the value of acceleration due to gravity, 9.8. This means that

9.8 = v²/1.9

v² = 1.9 * 9.8

v² = 18.62

v = √18.62

v = 4.32 m/s

This means that, the minimum speed the cup must have so as not to get wet or any spill is 4.32 m/s

6 0
2 years ago
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