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REY [17]
3 years ago
10

A plane is flying east at 115 m/s. The wind accelerates it at 2.88 m/s^2 directly northwest. After 25.0 seconds what is the dire

ction of the velocity of the plane?
Physics
1 answer:
Furkat [3]3 years ago
7 0

Answer:

38.5° north of east

Explanation:

Directly northwest is 135° from east.  If we say east is +x and north is +y, then the components of the acceleration are:

ax = 2.88 cos 135° ≈ -2.036 m/s²

ay = 2.88 sin 135° ≈ 2.036 m/s²

And the initial velocity is:

v₀x = 115 m/s

v₀y = 0 m/s

After, 25.0 seconds, the final velocity components are:

vx = at + v₀x

vx = (-2.036)(25.0) + 115

vx ≈ 64.1 m/s

vy = at + v₀y

vy = (2.036)(25.0) + 0

vy ≈ 50.9 m/s

The plane's direction is:

θ = atan(vy / vx)

θ = atan(50.9 / 64.1)

θ ≈ 38.5°

The direction of the plane's velocity is approximately 38.5° north of east.

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This question involves the concepts of the law of conservation of energy, kinetic energy, and potential energy.

The height of the hill is "166.76 m".

<h3>LAW OF CONSERVATION OF ENERGY:</h3>

According to the law of conservation of energy at the highest point of the roller coaster ride, that is, the hill, the whole (maximum) kinetic energy of the roller coaster is converted into its potential energy:

Maximum\ Kinetic\ Energy\ Lost = Maximum\ Potential\ Energy\ Gain\\\\&#10;\frac{1}{2}mv_{max}^2=mgh\\\\&#10;h=\frac{v_{max}^2}{2g}

where,

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Therefore,

h=\frac{(57.2\ m/s)^2}{2(9.81\ m/s^2)}\\\\&#10;

<u>h = 166.76 m</u>

Learn more about the law of conservation of energy here:

brainly.com/question/101125

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3 years ago
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Hi! Let me help you!
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