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3 years ago

Need help with two physics questions!

Physics

1 answer:

nikitadnepr [17]3 years ago

8 0

1) The north component of the airplane velocity is 260 km/h.

2) The direction of the plane is $24^{\circ}$ north of east.

Explanation:

In this problem, we have to resolve the velocity vector into its components.

Taking east as positive x-direction and north as positive y-direction, the components of the velocity along the two directions are given by:

$v_x = v cos \theta$

$v_y = v sin \theta$

where

v is the magnitude of the velocity

$\theta$ is the angle between the direction of the velocity and the positive x-axis (the east direction)

For the airplane in this problem,

v = 750 km/h

$\theta=20^{\circ}$

So, the two components are

$v_x = (750)(cos 20)=704.8 km/h$

$v_y = (750)(sin 20)=256.5 km/h$

So, the component in the north direction is 256.5 km/h, so approximately 260 km/h.

In this problem, we have to use vector addition.

In fact, the motion of the plane consists of two displacements:

- A first displacement of 220 km in the east direction

- A second displacement of 100 km in the north direction

Using the same convention of the same problem (x = east and y = north), we can write

$d_x = 220 km$

$d_y = 100 km$

Since the two vectors are perpendicular to each other, we can find their magnitude using Pythagorean's theorem:

$d=\sqrt{d_x^2+d_y^2}=\sqrt{(220)^2+(100)^2}=241.7 km$

And the direction is given by

$\theta=tan^{-1}(\frac{d_y}{d_x})=tan^{-1}(\frac{100}{220})=24^{\circ}$ north of east.

Learn more about vectors here:

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