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2 years ago

3.) An engineer is designing the runway for an airport. Of the planes that will use the airport,

Physics

2 answers:

valentinak56 [21]2 years ago

8 0

Answer:

704

Explanation:

scoray [572]2 years ago

5 0

Answer: 704

Explanation:Vi = 0 m/s

vf = 65 m/s

a = 3 m/s2

d = ??

vf2 = vi2 + 2*a*d

(65 m/s)2 = (0 m/s)2 + 2*(3 m/s2)*d

4225 m2/s2 = (0 m/s)2 + (6 m/s2)*d

(4225 m 2/m2)/(6 m/s2) = d

d = 704 m

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What two forces are there when you skydive

dusya [7]

Gravity is pulling you down and friction is slowing you down so you don't plummet to the ground at super high speeds.

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3 years ago

Read 2 more answers

A student walks 350 m [S], then 400 m [E20°N], and finally 550 m [N10°W]. Using the component method, find the resultant (total)

levacccp [35]

In component form, the displacement vectors become

• 350 m [S] ==> (0, -350) m

• 400 m [E 20° N] ==> (400 cos(20°), 400 sin(20°)) m

(which I interpret to mean 20° north of east]

• 550 m [N 10° W] ==> (550 cos(100°), 550 sin(100°)) m

Then the student's total displacement is the sum of these:

(0 + 400 cos(20°) + 550 cos(100°), -350 + 400 sin(20°) + 550 sin(100°)) m

≈ (280.371, 328.452) m

which leaves the student a distance of about 431.8 m from their starting point in a direction of around arctan(328.452/280.371) ≈ 50° from the horizontal, i.e. approximately 431.8 m [E 50° N].

5 0

2 years ago

GIVING BRAINLIEST PLEASE HELP!!

Marrrta [24]

The answer is D hope it helps

8 0

3 years ago

If two children, with masses of 16 kg and 24 kg , sit in seats opposite one another, what is the moment of inertia about the rot

Elena-2011 [213]

Answer:

The moment of inertia about the rotation axis is 117.45 kg-m²

Explanation:

Given that,

Mass of one child = 16 kg

Mass of second child = 24 kg

Suppose a playground toy has two seats, each 6.1 kg, attached to very light rods of length r = 1.5 m.

We need to calculate the moment of inertia

Using formula of moment of inertia

$I=I_{1}+I_{2}$

$I=(m+m_{1})\times r^2+(m+m_{2})\times r^2$

m = mass of seat

m₁ =mass of one child

m₂ = mass of second child

r = radius of rod

Put the value into the formula

$I=(16+6.1)\times(1.5)^2+(24+6.1)\times(1.5)^2$

$I=117.45\ kg-m^2$

Hence, The moment of inertia about the rotation axis is 117.45 kg-m²

8 0

3 years ago

Assume that an intercontinental ballistic missile goes from rest to a suborbital speed of 6.50 km/s in 60.0 s (the actual speed

olasank [31]

Answer:

Average acceleration is $(11.05)g\ m/s^2$

Explanation:

It is given that,

Initial velocity, u = 0

Final velocity, v = 6.5 km/s = 6500 m/s

Time taken, t = 60 s

Acceleration, $a=\dfrac{v-u}{t}$

$a=\dfrac{v}{t}$

$a=\dfrac{6500\ m/s}{60}$

$a=108.33\ m/s^2$

Since, $g=9.8\ m/s^2$

So, $a=(11.05)g\ m/s^2$

So, the angular acceleration of the missile is $(11.05)g\ m/s^2$ . Hence, this is the required solution.

4 0

2 years ago