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ddd [48]
2 years ago
5

A plane accelerates to a velocity of 240 m/s in 11 s by which time it had traveled 1,400 m down the runway what were it average

and initial velocities
Physics
1 answer:
Alex Ar [27]2 years ago
5 0

The average velocity is 127.27 m/s while the initial velocity is 14.55 m/s

According to Newton's law of motion:

s = [(v + u)/2]t

Where s is the distance moved = 1400 m, v is the final velocity = 240 m/s, u is the initial velocity, t is the time taken = 11 s, hence:

1400 = [(240 + u)/2)11

2800 = (240 + u)11

u + 240 = 254.55

u = 14.55 m/s

The average velocity = (v + u)/2 = (240 + 14.55)/2 = 127.27 m/s

Therefore the average velocity is 127.27 m/s while the initial velocity is 14.55 m/s

Find out more at: brainly.com/question/13639113

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Explanation:

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The model shows sunlight reaching Earth. Four points are shown on different parts of earth.
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2 years ago
The change in momentum experienced by a object is equivalent to the...
Papessa [141]

Answer: C (impulse acting on the object)

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From Newtons II law

              F = m. a

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8 0
3 years ago
Two circus performers rehearse a trick in which a ball and a dart collide. Horatio stands on a platform 9.8 m above the ground a
xenn [34]

Answer:

Time = 0.55 s

Height = 8.3 m

Explanation:

The ball is dropped and therefore has an initial velocity of 0. Its acceleration, g, is directed downward in the same direction as its displacement, h_b.

The dart is thrown up in which case acceleration, g, acts downward in an opposite direction to its displacement, h_d. Both collide after travelling for a time period, t. Let the height of the dart from the ground at collision be h_d and the distance travelled by the ball measured from the top be h_b.

It follows that h_d+h_b=9.8.

Applying the equation of motion to each body (h = v_0t + 0.5at^2),

Ball:

h_b=0\times t + 0.5\times 9.8t^2 (since v_{b0} =0.)

h_b=4.9t^2

Dart:

h_d=17.8\times t - 0.5\times9.8t^2 (the acceleration is opposite to the displacement, hence the negative sign)

h_d=17.8\times t - 4.9t^2

But

h_b+h_d =9.8

17.8\times t - 4.9t^2+4.9t^2 =9.8

17.8\times t = 9.8

t = 0.55

The height of the collision is the height of the dart above the ground, h_d.

h_d=17.8\times t - 4.9t^2

h_d=17.8\times 0.55 - 4.9\times(0.55)^2

h_d=9.79 - 1.48225

h_d=8.3

8 0
3 years ago
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