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Alina [70]
3 years ago
11

Features that describe motion of an object

Physics
1 answer:
irakobra [83]3 years ago
7 0
 Since motion can be discribed as distance, acceleration, time, speed, and displacement, then I believe the features that describe motion of an object would be all of these.<span> </span>
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Yellow-green light has a wavelength of 560 nm. What is its frequency?
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5.4 x 1014Hz
wavelength x frequency = the speed of light
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The density of a substance can be found with the help of:
Bess [88]
Mass (m) and volume (v)
the equation is d = m/v
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3 years ago
Equations E = 1 2πε0 qd z3 and E = 1 2πε0 P z3 are approximations of the magnitude of the electric field of an electric dipole,
tamaranim1 [39]

Answer:

The ratio of E_{app} and E_{act} is 0.9754

Explanation:

Given that,

Distance z = 4.50 d

First equation is

E_{act}=\dfrac{qd}{2\pi\epsilon_{0}\times z^3}

E_{act}=\dfrac{Pz}{2\pi\epsilon_{0}\times (z^2-\dfrac{d^2}{4})^2}

Second equation is

E_{app}=\dfrac{P}{2\pi\epsilon_{0}\times z^3}

We need to calculate the ratio of E_{act} and E_{app}

Using formula

\dfrac{E_{app}}{E_{act}}=\dfrac{\dfrac{P}{2\pi\epsilon_{0}\times z^3}}{\dfrac{Pz}{2\pi\epsilon_{0}\times (z^2-\dfrac{d^2}{4})^2}}

\dfrac{E_{app}}{E_{act}}=\dfrac{(z^2-\dfrac{d^2}{4})^2}{z^3(z)}

Put the value into the formula

\dfrac{E_{app}}{E_{act}}=\dfrac{((4.50d)^2-\dfrac{d^2}{4})^2}{(4.50d)^3\times4.50d}

\dfrac{E_{app}}{E_{act}}=0.9754

Hence, The ratio of E_{app} and E_{act} is 0.9754

8 0
2 years ago
Determine the distance between a newly discovered planet and its single moon if the orbital period of the moon is 1.2 Earth days
Vinil7 [7]

Answer:

The distance is r = 55430496 \  m  

Explanation:

From the question we are told that

   The period of the moon T =  1.2 days = 1.2 * 24 * 3600 = 103680 \  s

    The mass of the planet is  m_p =  9.38*10^{24} kg

Generally the period of the moon is mathematically represented as

       T  =  2 *  \pi * \sqrt{ \frac{r^3 }{ G * m_p } }

Here G is the gravitational constant with value

        G  =  6.67 *10^{-11} \  N \cdot m^2/kg^2

=>   T  =  2 *  \pi * \sqrt{ \frac{r^3 }{ G * m_p } }

=>   103680   =  2 *  3.142  * \sqrt{ \frac{r^3 }{ 6.67*10^{-11} * 9.38*10^{24} } }

=>    272218492.31 = \frac{r^3}{ 6.67 *10^{-11} * 9.38*10^{24}}

=>    r = \sqrt[3]{ 1.7031241*10^{23}}j

=>   r = 55430496 \  m        

8 0
3 years ago
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