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A Boeing 787 is initially moving down the runway at 6.0 m/s preparing for takeoff. The pilot pulls on the throttle so that the e

andrey2020 [161]

Answer:

39.26 s

Explanation:

From the question given above, the following data were obtainedb

Initial velocity (u) = 6 m/s

Acceleration (a) = 1.9 m/s²

Distance travelled (s) = 1700 m

Time (t) =?

Next, we shall determine the final velocity of the plane. This can be obtained as follow:

Initial velocity (u) = 6 m/s

Acceleration (a) = 1.9 m/s²

Distance travelled (s) = 1700 m

Final velocity (v) =?

v² = u² + 2as

v² = 6² + (2 × 1.9 × 1700)

v² = 36 + 6460

v² = 6496

Take the square root of both side

v = √6496

v = 80.6 m/s

Finally, we shall determine the time taken before the plane lifts off. This can be obtained as follow:

Initial velocity (u) = 6 m/s

Acceleration (a) = 1.9 m/s²

Final velocity (v) = 80.6 m/s

Time (t) =?

v = u + at

80.6 = 6 + 1.9t

Collect like terms

80.6 – 6 = 1.9t

74.6 = 1.9t

Divide both side by 1.9

t = 74.6 / 1.9

t = 39.26 s

This, it will take 39.26 s before the plane lifts off.

3 0

3 years ago

A ball is batted straight up into the air and reaches a maxium height 65.6 m (a) How long did it take to reach this height? (b)

kondaur [170]

Answer:

a) 3.65 seconds

b) 35.87 m/s

Explanation:

s = Displacement = 65.6 m

u = Initial velocity

v = Final velocity

t = Time taken

a = Acceleration due to gravity = 9.81 m/s² (downward direction is taken as positive and upward is taken as negative)

b) Equation of motion

$v^2-u^2=2as\\\Rightarrow 0^2-u^2=2\times -9.81\times 65.6\\\Rightarrow u=\sqrt{2\times 9.81\times 65.6}\\\Rightarrow u=35.87\ m/s$

Initial pop up velocity is 35.87 m/s

a)

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-35.87}{-9.81}\\\Rightarrow t=3.65\ s$

It took 3.65 seconds to reach this height

6 0

3 years ago

When you are in the way of a moving object and a collision is sure to occur, you are better off decreasing its momentum over …?

BlackZzzverrR [31]

<span>When you are in the way of a moving object and a collision is sure to occur, you are better off decreasing its momentum over time. Which is nothing but force.
mv/t = m.a = F

Fill the blank as: "Time"

Hope this helps!</span>

7 0

3 years ago

Jan ran 4 miles north in 28 minutes. What was Jan's average velocity?

fenix001 [56]

Answer:

3.83 m/s

Explanation:

Given that,

Distance covered by Jan, d = 4 miles

1 mile = 1609.34 m

4 miles = 6437.38 m

Time, t = 28 minutes = 1680 s

Jan's average speed,

v = d/t

$v=\dfrac{6437.38\ \text{m}}{1680\ \text{s}}\\\\v=3.83\ \text{m/s}$

Hence, the average velocity of Jan is 3.83 m/s.

6 0

3 years ago

There are two blocks: one large one initially at rest, and a smaller one, initially moving to the right withsome speed. The smal

KATRIN_1 [288]

Answer:

Explanation:

Let the initial velocity of small block be v .

by applying conservation of momentum we can find velocity of common mass

25 v = 75 V , V is velocity of common mass after collision.

V = v / 3

For reaching the height we shall apply conservation of mechanical energy

1/2 m v² = mgh

1/2 x 75 x V² = 75 x g x 10

V² = 2g x 10

v² / 9 = 2 x 9.8 x 10

v² = 9 x 2 x 9.8 x 10

v = 42 m /s

small block must have velocity of 42 m /s .

Impulse by small block on large block

= change in momentum of large block

= 75 x V

= 75 x 42 / 3

= 1050 Ns.

6 0

3 years ago