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avanturin [10]
3 years ago
10

3. An airplane is heading north with an airspeed of 325 m/s with a wind from the east at 55.0 m/s. What is the airplane's veloci

ty relative to the ground? Is the airplane's ground speed faster or slowerthan its air speed? Why?​
Physics
1 answer:
liraira [26]3 years ago
3 0

Let <em>a</em> denote the airplane's velocity in the air, <em>g</em> its velocity on the ground, and <em>w</em> the velocity of the wind. (Note that these are vectors.) Then

<em>a</em> = <em>g</em> + <em>w</em>

and we're given

<em>a</em> = (325 m/s) <em>j</em>

<em>w</em> = (55.0 m/s) <em>i</em>

Then

<em>g</em> = - (55.0 m/s) <em>i</em> + (325 m/s) <em>j</em>

The ground speed is the magnitude of this vector:

||<em>g</em>|| = √[ (-55.0 m/s)² + (325 m/s)² ] ≈ 330. m/s

which is faster than the air speed, which is ||<em>a</em>|| = 325 m/s.

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An elevator supported by a single cable descends a shaft at a constant speed. The only forces acting on the elevator are the ten
Gala2k [10]

Answer:

E) The net work done by the two forces is zero joules.

Explanation:

As the elevator is moving at a constant speed, it is the same as being stationary. The tension in the cable perfectly balances the gravitational force. Work done by gravity = mgh

Work done by the tension = -Th

where, T = tension force. The negative sign comes as the direction of T is opposite to the direction of g. As the elevator moves at a constant speed, the magnitude of the two forces must be equal. Therefore, the total work done,

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A rocket is launched from Earth (mass ME, radius RE) with velocity v° and reaches radial distance r=6RE with velocity v°/10. Exp
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The velocity, expressed in terms of  ME, RE is given as V_0 = \sqrt{98.99R_E}.

The maximum height that the rocket could reach if launched vertically is H = (¹/₂v₀²)/g.

<h3>Maximum height of the rocket</h3>

The maximum height reached by the rocket can be modeled using conservation of energy as shown below;

P.Ei + K.Ei = K.Ef + P.Ef

M_EgR_E + \frac{1}{2} M_EV_0^2= \frac{1}{2} M_E(\frac{V_o}{10} )^2+ M_Eg(6R_E)\\\\\frac{1}{2} M_EV_0^2 - \frac{1}{2} M_E(\frac{V_o}{10} )^2 = M_Eg(6R_E) - M_EgR_E\\\\0.495M_EV_0^2= 5gM_ER_E\\\\0.495V_0^2= 5gR_E\\\\V_0 = \sqrt{\frac{5gR_E}{0.495} } \\\\V_0 = \sqrt{98.99R_E}

<h3>Maximum height when it is launched vertically</h3>

P.E = K.E

mgH = ¹/₂mv²

H = (¹/₂v₀²)/g

Learn more about conservation of energy here: brainly.com/question/166559

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Pick the correct word to fill up the blank in below question: velocity, combustion, acceleration, gravitation, force
scZoUnD [109]

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The altitude of the International Space Station ttt minutes after its perigee (closest point), in kilometers, is given by \qquad
Dmitriy789 [7]

Answer:

T = 92.8 min

Explanation:

Given:

The altitude of the International Space Station t minutes after its perigee (closest point), in kilometers, is given by:

                               A(t) = 415 - sin(\frac{2*\pi (t+23.2)}{92.8})

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Hence,                     T = 92.8 min

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