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OleMash [197]
3 years ago
7

Which statement best describes perigee?

Physics
1 answer:
Alex17521 [72]3 years ago
6 0

Whenever you see "-gee" in the description of a point in an orbit,
you know it's talking about an orbit of the Earth.  You see the same
piece of a word in "geology" and "geography".  "Geometry" began
as the study of measuring places on the Earth, so that you and your
neighbor could agree on where your field ends and his begins, and
if you wanted to buy part of his field from him, the two of you could
go outside, do some measurements, and agree on what area you're
paying him for.

"Perigee" and "apogee" are the points in the orbit of the Moon, or a
TV satellite, or the International Space Station, where the orbiting body
is nearest or farthest from the Earth.  "Perigee" is the lowest/nearest point. 
"Apogee" is the highest/farthest point.

If the description has "-helion" in it instead of "-gee", then it's talking about
an orbit around the sun, like points in the Earth's orbit.  The "-helion" comes
from the Greek word "Helios" for the Sun.
Earth is at perihelion during the first few days of January, and at aphelion
during the first few days of July.  (That's right ... nearest to the sun in January,
and farthest from the sun in July.)

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DNA of two species is one of the best evidence to tell us how closely related the two species are.
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3 years ago
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An electromagnetic wave is transporting energy in the negative y direction. At one point and one instant the magnetic field is i
natali 33 [55]

Answer:. Option c

Explanation: the speed of an electromagnetic wave is simply the vector product of the magnetic field and the electric field.

The direction of the velocity is the direction of the electromagnetic wave.

The wave is already moving towards the negative y axis (-j) and the magnetic field is already pointing towards the positive x axis (i)

From cross product of unit vectors

i × j = k

i × k = - j

With the second identity, we can see that the electric field will be pointing towards the positive of the x axis (k).

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4 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

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3 years ago
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Tema [17]
No, I heavier object will fall much faster than something lighter than it. This is because it’s more dense and hard so it can cut through the air particles quicker than a lighter object which takes longer to cut through the air and fall

Example:

A rock vs a feather
The rock will fall quicker because it’s more dense and falls straight down and the feather will be slower because it flows slowly down through the air particles
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