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galina1969 [7]
3 years ago
6

I need help pls. Solve 1+cos 30º/1-sin 30º. Plz help me​

Physics
1 answer:
Maurinko [17]3 years ago
6 0

Answer:

\huge \boxed{ \boxed{2 +  \sqrt{3}}}

Explanation:

<h3>to understand this</h3><h3>you need to know about:</h3>
  • trigonometry
  • PEMDAS
<h3>given:</h3>
  • \frac{1 +  \cos( {30}^{o} ) }{1 -  \sin( {30}^{o} ) }
<h3>tips and formulas:</h3>
  • cos(30°)=\frac{\sqrt{3}}{2}
  • \sin( {30}^{o} )  =  \frac{1}{2}
<h3>let's solve:</h3>
  1. \sf sustitute \: the \: value \: of \:  \sin( {30}^{o} ) and \cos( {30}^{o} )  \\  \sf \frac{1 +   \frac{ \sqrt{3} }{2}}{1 -  \frac{1}{2} }
  2. \sf simplify :  \\  \frac{ \frac{2 +  \sqrt{3} }{2} }{ \frac{1}{2} }
  3. \sf simplify(fraction) :  \\  \frac{2 +  \sqrt{3} }{2}  \div  \frac{1}{2}  \\ \frac{2 +  \sqrt{3} }{2} \times 2 \\  \therefore 2 +  \sqrt{3}

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Calculate the orbital period of a dwarf planet found to have a semimajor axis of a = 4.0x 10^12 meters in seconds and years.
padilas [110]

Explanation:

We have,

Semimajor axis is 4\times 10^{12}\ m

It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :

T^2=\dfrac{4\pi ^2}{GM}a^3

G is universal gravitational constant

M is solar mass

Plugging all the values,

T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s

Since,

1\ s=3.17\times 10^{-8}\ \text{years}\\\\4.37\times 10^9\ s=4.37\cdot10^{9}\cdot3.17\cdot10^{-8}\\\\4.37\times 10^9\ s=138.52\ \text{years}

So, the orbital period of a dwarf planet is 138.52 years.

3 0
3 years ago
which of the glass lenses above, when placed in air, will cause rays of light (parallel to the central axis) to converge?
777dan777 [17]

Convex lenses when placed in the air, will cause rays of light (parallel to the central axis) to converge.

Converging lenses, commonly referred to as convex lenses, have thicker centers and narrower upper and lower margins. The edges are outwardly curled. This lens has the ability to concentrate a beam of parallel light rays coming from the outside onto a spot on the opposite side of the lens.

The image created is referred to be a genuine image when it is inverted relative to the object. On a screen, this kind of image can be recorded. When the object is positioned at a point farther than one focal length from the lens, a converging lens creates a true image.

A virtual image is one that cannot be produced on a screen and is formed when the image is upright in relation to the object. When an item is positioned within one focal length of a converging lens, a virtual image is created. It creates an enlarged image of the object on the same side of the lens as the image. It serves as a magnifier.

Learn more about the convex lens here:

brainly.com/question/12847657

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3 0
1 year ago
A particle is moving with (SHM) of period 8.0s and amplitude5.0m
nadezda [96]

Answer:

velocity(x)=15\,\frac{\pi}{4}\,cos(\frac{\pi}{4}x)

Max speed = \frac{15\, \pi}{4} \,\, \frac{m}{s}

Max acceleration = \frac{15\,\pi^2}{16} \,\,\frac{m}{s^2}

Explanation:

Given the description of period and amplitude, the SHM could be described by:

f(x)=5\,sin(\frac{\pi}{4}x)

and its angular velocity can be calculated doing the derivative:

f(x)=5\, \,sin(\frac{\pi}{4}x)\\f'(x)=5\,\frac{\pi}{4}\,cos(\frac{\pi}{4}x)

And therefore, the tangential velocity is calculated by multiplying this expression times the radius of the movement (3 m):

velocity(x)=15\,\frac{\pi}{4}\,cos(\frac{\pi}{4}x)  and is given in m/s.

Then the maximum speed is obtained when the cosine function becomes "1", and that gives:

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The acceleration is found from the derivative of the velocity expression, and therefore given by:

acceleraton(x)=-15\,\frac{\pi^2}{16}\,sin(\frac{\pi}{4}x)

and the maximum of the function will be obtained when the sine expression becomes "-1", which will render:

Max acceleration = \frac{15\,\pi^2}{16} \,\,\frac{m}{s^2}

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3 years ago
Over a period of operation, the useful work output of the fluorescent bulb was
Nadya [2.5K]

Answer:

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Un camion de envios se encuentra detenido en una señal de pare, permitiendo que pase una ambulancia. Inicia su recorrido y al ca
Nesterboy [21]

Answer:

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u = Velocidad inicial = 0

t = Tiempo empleado = 15 s

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De las ecuaciones cinemáticas tenemos

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