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1 year ago

Find the x-component of this

Physics

2 answers:

Mashcka [7]1 year ago

8 0

The horizontal or x component of the vector is 53.5 m.

<h3>What is the x component?</h3>

We know that a vector quantity must posses both magnitude and direction. In this case, of a vector is inclined at an angle then the vector could have a vertical and a horizontal component. Recall that vector may have have;

- A vertical component

- A horizontal component

Recall that when we resolve a vector, we are able to obtain the effective component of the vector in the given direction. It could be vertical or horizontal.

The horizontal component of the vector is V cos θ where V is the numerical value of the vector.

Thus the horizontal component is;

55.1 m Cos 13.9 = 53.5 m

Learn more about components of a vector:brainly.com/question/4179238

#SPJ1

mash [69]1 year ago

4 0

Answer:

13.24

Explanation:

Fx=Fcos(theta)

but remember theta is measured from the x axis, so in this case it makes theta 76.1 degrees and not 13.9.

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siniylev [52]

Answer:

$v = 4.44 \times 10^5 m/s$

Explanation:

By Einstein's Equation of photoelectric effect we know that

$h\nu = W + \frac{1}{2}mv^2$

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$h\nu$ = energy of the photons incident on the metal

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$W = \frac{hc}{351 nm}$

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$\frac{1242 eV-nm}{303 nm} = 3.54 eV + \frac{1}{2}(9.1 \times 10^{-31})v^2$

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3 years ago

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

Vladimir79 [104]

Answer:

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

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.

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