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kakasveta [241]
4 years ago
11

What are the historical solar system models. Differences

Physics
1 answer:
Rasek [7]4 years ago
7 0
Aristarchus(310-230 BC)
Aristarchus was a Greek Astronomer and Mathematician. He proposed the heliocentric model of the Solar System. He attempted to measure the distances from the moon to the sun.
Aristotle (334-322 BC)
<span>Aristotle was a Greek Philosopher. He was the first person to propose this historical model. He thought the Earth was in the center of all celestial beings. His idea was that the Earth was in the middle of the sun, moon, stars, and other planets. </span>
Ptolemy(100-170 CE)
<span>Ptolemy was a astronomer and a geographer and a mathematician! He proposed the geocentric model. He believed that a celestial body could go at a constant speed in a perfect circle. It was called the "wheels-on-wheels" system. It had better observations than the model that had came before his. His geocentric model was used for the next 14 centuries! </span>
Kepler(1571-1630)
<span>German astronomer and mathematician, Johannes Kepler, studied very hard and closely, and he found out that the planets did not fit the observations that were made by previous explorers. He found one that fit the best out of them all! He found three principles, called Kepler's Laws. </span>
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58.5 million excess electrons are inside a closed surface. What is the net electric flux through the surface?
Anni [7]
The biggest thing you're doing wrong is ignoring the units
when you're working with the quantities.

Now let's look at the rest of the problem:

The formula you used is correct:

           Net flux through the surface = (net charge inside) / ε₀

and          ε₀ = 8.85 x 10⁻¹² farad/meter.

What's the net charge inside the surface in this problem ?

It's    (5.85 x 10⁷ electrons) x (the charge on each electron)

     =  (5.85 x 10⁷ electrons) x (-1.6 x 10⁻¹⁹ coulomb/electron)

     =      -9.36 x 10⁻¹² coulomb .   

Finally,      (net charge inside) / ε₀

             =  (-9.36 x 10⁻¹² coulomb) / (8.85 x 10⁻¹² farad/meter)

             =        -1.058  newton-m²/coulomb .

The sign and the significant figures in your answer are correct, so
we can see that you know what you're doing.  The only thing left is
the order of magnitude.  You most likely took one of the negative
exponents and made it positive.  You got an answer that's 10²² too
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8 0
3 years ago
If the magnitude of the moment of F about line CD is 57 N·m, determine the magnitude of F.If the magnitude of the moment of F ab
bazaltina [42]

Answer:

F_ab = 260.17 N

Explanation:

Given:

- Moment of force F about CD, (M)_cd = 57 Nm

Find:

- First we will write down the position vectors of points A, B , C , D:

- We will take left and bottom most corner of cube to be the origin.

- The unit vectors i , j , k are along vertical planes and outside the plane, respectively.

- The position vectors wrt to the origin are:

                             Point A = 0.2 k

                             Point B = 0.4 i + 0.2 j

                             Point C = 0.2 j + 0.4 k

                             Point D = 0.4 i + 0.4 k

- Now we will determine the Force vector F_ab along vector AB.

                             vec (AB) = B - A = 0.4 i + 0.2 j - 0.2 k

                             unit (AB) = 0.4 i + 0.2 j - 0.2 k  / sqrt ( 0.4^2 + 2*0.2^2)

                                            = [5 / sqrt(6)] * ( 0.4 i + 0.2 j - 0.2 k )

Hence,

                              vec(F_ab) = Fab*[5 / sqrt(6)] * ( 0.4 i + 0.2 j - 0.2 k )

- Now, form a unit vector along the line CD:

                             vec(CD) = D - C = 0.4 i - 0.2 j

                             unit (CD) = 0.4 i - 0.2 j / sqrt ( 0.4^2 + 0.2^2)

                                           = [sqrt(5)]*(0.4 i - 0.2 j)

- Now select a point on line CD, lets say C. Find the moment arm from line of action of force along AB and line CD. Hence, vector AC is:

                               vec(AC) =r_ac = C - A = 0.2 j + 0.2 k

- Now the moment about a line CD due to force is:

                              (M)_cd = unit(CD) . ( r_ac x vec(F_ab) )

The cross product of r_ac and vec(F_ab) is as follows:

                               (M)_c =  ( r_ac x vec(F_ab) ) :

                                \left[\begin{array}{ccc}i&j&k\\0&0.2&0.2\\0.8165&0.40824&-0.40824\end{array}\right]

                              (M)_c =  F_ab[- sqrt(6)/15 i + sqrt(6)/15 j - sqrt(6)/15 k]

The dot product of (M)_c and unit (CD)  is as follows:

                              (M)_cd = unit(CD) . (M)_c :

 (M)_cd = F_ab[- sqrt(6)/15 i + sqrt(6)/15 j - sqrt(6)/15 k] .  [sqrt(5)]*(0.4 i - 0.2 j)

                              (M)_cd = F_ab*(sqrt(30) / 25)

- The given magnitude of the moment is (M)_cd. Calculate F_ab:

                               57 = F_ab*(sqrt(30) / 25)  

                              F_ab = 260.17 N

7 0
4 years ago
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