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3 years ago

Find the magnitude and direction for 101m,60.0,85.0m

Physics

1 answer:

Crank3 years ago

4 0

Answer:

magnitude = 161.3m, ∅ = 32.9°

Explanation:

Vector addition always works the same. Add two vectors by adding their respective components.

vector A: $\left[\begin{array}{c}85.0&0.0\end{array}\right]$

vector B: $101.0\left[\begin{array}{c} cos60.0&sin 60.0\end{array}\right] =\left[\begin{array}{c}50.5&87.5\end{array}\right]$

Adding vector A and B: $\left[\begin{array}{c}85.0&0.0\end{array}\right] +\left[\begin{array}{c}50.5&87.5\end{array}\right] = \left[\begin{array}{c}135.5&87.5\end{array}\right]$

The magnitude of any vector $\left[\begin{array}{c}a&b\end{array}\right]$ is given by the Pythagorean theorem:

$magnitude = \sqrt{a^2+b^2}$

In the case of the vector A+B:

$magnitude = \sqrt{135.5^2+87.5^2}$

The angle ∅ of the vector can by found by using trigonometric functions:

For instance, the angle ∅ for a vector $\left[\begin{array}{c}a&b\end{array}\right]$ is given by the equation:

$tan\phi= \frac{b}{a}$

The direction ∅ can be found by solving the trigonometric function.

In the example of vector A+B:

$tan\phi = \frac{87.5}{135.5}$

Solving for ∅:

$\phi = tan^{-1} (\frac{87.5}{135.5})=32.9$

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3 years ago

Which is an example of transforming potential energy to kinetic energy? Select two options.

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Complete Question:

Which is an example of transforming potential energy to kinetic energy? Select two options.

changing thermal energy to electrical energy

changing chemical energy to thermal energy

changing nuclear energy to radiant energy

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changing mechanical energy to chemical energy

Correct Answer:

The examples of transforming potential energy to kinetic energy are changing chemical energy to thermal energy and changing nuclear energy to radiant energy.

Explanation:

As stated by the conservation of energy law, any form of energy is usually transferred to another form. The basic kinds of energy is potential and kinetic energy. Potential energy is the energy stored for the objects at rest and kinetic energy is the energy utilized by the objects for motion.

So in the given options, chemical energy is the energy stored by the chemical bonds to make a stable compound and that energy is converted to thermal energy when the bonds get broken. So the stored energy or the energy required to keep the bonds intact is chemical energy and it is thus a form of potential energy.

And when these bonds get broken, the electrons use the thermal energy released by this breakage as their kinetic energy. So one form of transforming potential energy to kinetic energy is by changing chemical energy to thermal energy.

Similarly, the nuclear energy is exhibited by the elementary particles in an atom. So it is similar to potential energy and the radiant energy is released whenever there is an excitation. So the radiant energy will be similar to kinetic energy.

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3 years ago

A rocket is launched at an angle of 53.0° above the horizontal with an initial speed of 103 m/s. The rocket moves for 3.00 s alo

Serggg [28]

Before the engines fail $(0\le t\le3.00\,\rm s)$ , the rocket's horizontal and vertical position in the air are

$x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t^2$

$y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t^2$

and its velocity vector has components

$v_x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t$

$v_y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t$

After $t=3.00\,\rm s$ , its position is

$x=273\,\rm m$

$y=362\,\rm m$

and the rocket's velocity vector has horizontal and vertical components

$v_x=120\,\frac{\rm m}{\rm s}$

$v_y=159\,\frac{\rm m}{\rm s}$

After the engine failure $(t>3.00\,\rm s)$ , the rocket is in freefall and its position is given by

$x=273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)t$

$y=362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

and its velocity vector's components are

$v_x=120\,\frac{\rm m}{\rm s}$

$v_y=159\,\frac{\rm m}{\rm s}-gt$

where we take $g=9.80\,\frac{\rm m}{\mathrm s^2}$ .

a. The maximum altitude occurs at the point during which $v_y=0$ :

$159\,\frac{\rm m}{\rm s}-gt=0\implies t=16.2\,\rm s$

At this point, the rocket has an altitude of

$362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)(16.2\,\rm s)-\dfrac g2(16.2\,\rm s)^2=1650\,\rm m$

b. The rocket will eventually fall to the ground at some point after its engines fail. We solve $y=0$ for $t$ , then add 3 seconds to this time:

$362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=34.6\,\rm s$

So the rocket stays in the air for a total of $37.6\,\rm s$ .

c. After the engine failure, the rocket traveled for about 34.6 seconds, so we evalute $x$ for this time $t$ :

$273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)(34.6\,\rm s)=4410\,\rm m$

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3 years ago

If a light is moved twice (2x) as far from a surface, the area the light covers is ___ as big.

Ksenya-84 [330]

Answer:

twice

Explanation:

From magnification = height of image / height of object

Distance of image/ distance of object = magnification

If the distance and height of the object represents the initial light distance and the exposed surface respectively.

And similarly the distance and height of the image represents the final light distance and the exposed surface respectively.

Hence the new image exposure would be twice as large.

If we use the formula our point of investigation is Height of image,

H2= D2/D1× H1

H2 = 2D2/D1 × H1

H2 = 2H1

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3 years ago