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3 years ago

- An experienced pilot can keep an engineless glider aircraft aloft for hours by

Physics

1 answer:

Annette [7]3 years ago

6 0

Answer:

Flift=5.00×10³ N

Fhorizontal=0.50×10³ N

Fg=4.80×10³ N

Explanation:

A free-body diagram of the glider.

pretend this is a number line

| Flift

|---->Fhorizontal

| Fg

∨

hope this helps! :)

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A ship sailing in the Gulf Stream is heading 25.0º west of north at a speed of 4.00 m/s relative to the water. Its velocity rela

ASHA 777 [7]

Answer:

Velocity of Gulf Stream= $34.5^\circ$ west of north at a speed of $2.03\ \rm m/s$

Explanation:

Given

speed of ship relative to Gulf stream= $25^\circ$ west of north at a speed of 4 m/s

speed of ship relative to earth= $5^\circ$ west of north at a speed of 4 .8 m/s

Let $V_S$ be the velocity of the ship relative to the earth and $V_{sg}$ be the velocity of the ship with respect to the Gulf stream

Let the west direction be negative x axis and north direction be positive x axis

Now

$V_{sg}=-4\sin25^\circ \vec i+4\cos25^\circ \vec j\\V_s-V_g=-4\sin25^\circ \vec i+4\cos25^\circ \vec j\\-4.8\sin5^\circ \vec i+4.8\cos5^\circ \vec j-V_g=-4\sin25^\circ \vec i+4\cos25^\circ \vec j\\Vg=-1.68\vec\ i+1.156\vec\ j\ \rm m/s\\$

magnitude of the velocity is given by

$V_g=\sqrt{(1.68^2+1.156^2)}\\V_g=2.03\ \rm m/s\\$

Let the velocity of gulf stream makes an angle $\theta$ with the positive y axis we have

$\tan\theta=\dfrac{1.156}{1.68}\\\theta=34.5^\circ$

Velocity of Gulf Stream $34.5^\circ$ west of north at a speed of $2.03\ \rm m/s$

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