Physics, 16.03.2020 19:21 carlybeavers50

1. A DC-10 jumbo jet maintains an airspeed of 550 mph in a southwesterly direction. The velocity of the jet stream is a constant 80 mph from the west. Find the actual speed and direction of the aircraft.

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The magnitude of the vectors is calculated by the Pythagorean theorem. the actual speed of the jet stream is 496.67 mph.

Velocity:

Given here,

Speed in southwest direction = 550 mph

Speed in west = 80 mph

Speed in southwest direction,

$\vec v_s_w = -550 \rm {\ cos (45)} \it i +\rm (-550 \rm sin (45 )\it j)\\\vec v_s_w = \rm -388 \it i - \rm 388 \it j\\$

Speed in west direction,

$\vec v_w = 80 i$

Add these values,

$\vec v = -5550 i - 550 j +80 i\\\\\vec v = -308.91 i - 388 j$

From the Pythagorean theorem,

$\vec v = \sqrt { -308.91 i - 388 j}\\\\\vec v = 496.67\rm \ mph$

Therefore, the actual speed of the jet stream is 496.67 mph.

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1. A DC-10 jumbo jet maintains an airspeed of 550 mph in a southwesterly direction. The velocity of

Answer from: ahmedeldyame

The magnitude of actual velocity is 496.67 mph and its direction is 51.54° with the x axis in the third quadrant.

Explanation:

Given:

Speed of jumbo jet in southwesterly direction (v_j) = 550 mph

Velocity of jet stream from west to east direction $(v_s)=80\ mph$

First let us draw a vectorial representation of the above velocity vectors.

Consider the south direction as negative y axis and west direction as negative x axis.

From the diagram,

The velocity of the jet can be represented as:

$\vec{v_j}=-550\cos(45)\vec{i}+(-550\sin(45)\vec{j} )\\\\\vec{v_j}=-388.91\vec{i}-388.91\vec{j}\ mph$

Similarly, the velocity of the stream is, $\vec{v_s}=80\vec{i}$

Now, the vector sum of the above two vectors gives the actual velocity of the aircraft. So, the resultant velocity is given as:

$\vec{v}=\vec{v_j}+\vec{v_s}\\\\\vec{v}=-388.91\vec{i}-388.91\vec{j}+80\vec{i}\\\\\vec{v}=(-388.91+80)\vec{i}-388.91\vec{j}\\\\\vec{v}=(-308.91)\vec{i}-388.91\vec{j}$

Now, magnitude is given as the square root of sum of the squares of the 'i' and 'j' components. So,

$|\vec{v}|=\sqrt{(-308.91)^2+(-388.91)^2}\\\\|\vec{v}|=496.67\ mph$

As the horizontal and vertical components of actual velocity negative, the resultant vector makes an angle $\theta$ with the x axis in the third quadrant.

The direction is given as:

$\theta=\tan^{-1}(\frac{v_y}{v_x})\\\\\theta=\tan^{-1}(\frac{-388.91}{-308.91})\\\\\theta=51.54\°(Third\ quadrant)$

Therefore, the magnitude of actual velocity is 496.67 mph and its direction is 51.54° with the x axis in the third quadrant.

Answer from: Quest

when insulating materials rub against each other, they may become electrically charged . electrons , which are negatively charged, may be 'rubbed off' one material and on to the other. the material that gains electrons becomes negatively charged. the material that loses electrons is left with a positive charge.

Answer from: Quest

probably not if net force is zero

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