Answer:
eccentrcity of orbit is 0.22
Explanation:
GIVEN DATA:
Initial velocity of satellite = 8333.3 m/s
distance from the sun is 600 km
radius of earth is 6378 km
as satellite is acting parallel to the earth therefore![\theta angle = 0](https://tex.z-dn.net/?f=%20%5Ctheta%20angle%20%3D%200)
and radial component of given velocity is zero
we have![h = r_o v_r_o = 6378+600 =6.97*10^6 m](https://tex.z-dn.net/?f=%20h%20%3D%20r_o%20v_r_o%20%3D%206378%2B600%20%3D6.97%2A10%5E6%20m)
h = 6.97*10^6 *8333.3 = 58.08*10^9 m^2/s
we know that
![\frac{1}{r} =\frac{GM}{h^2} \times ( i + \epsilon cos\theta)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Br%7D%20%3D%5Cfrac%7BGM%7D%7Bh%5E2%7D%20%5Ctimes%20%28%20i%20%2B%20%5Cepsilon%20cos%5Ctheta%29)
![GM = gr^2 = 9.81*(6.37*10^6)^2 = 398*10^{12} m^3/s](https://tex.z-dn.net/?f=GM%20%3D%20gr%5E2%20%3D%209.81%2A%286.37%2A10%5E6%29%5E2%20%3D%20398%2A10%5E%7B12%7D%20m%5E3%2Fs)
so
![\frac{1}{6.97*10^6} =\frac{398*10^{12}}{(58.08*10^9)^2} \times ( i + \epsilon cos0)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B6.97%2A10%5E6%7D%20%3D%5Cfrac%7B398%2A10%5E%7B12%7D%7D%7B%2858.08%2A10%5E9%29%5E2%7D%20%5Ctimes%20%28%20i%20%2B%20%5Cepsilon%20cos0%29)
solvingt for ![\epsilon)](https://tex.z-dn.net/?f=%20%5Cepsilon%29)
![\epsilon = 0.22)](https://tex.z-dn.net/?f=%5Cepsilon%20%3D%200.22%29)
therefore eccentrcity of orbit is 0.22
Technician B is right say that hard water potting i usually jut a Surface problem that can be wahed off.
What do you mean by Hard water?
The amount of dissolved calcium and magnesium in the water determines its hardness. Calcium and magnesium are the main dissolved minerals in hard water. The last time you washed your hands, you might have actually felt the effects of hard water.
What do you mean by acid rain?
Any type of precipitation that contains acidic elements, such as sulfuric or nitric acid, that falls to the ground from the atmosphere in wet or dry forms is referred to as acid rain, also known as acid deposition. Rain, snow, fog, hail, and even acidic dust can fall under this category.
Some plants are sensitive to excessive moisture around their root zone, so it may be necessary to increase drainage when growing plants in pots. Additionally, standing water at the bottom of the pot can cause root rot.
Many university agriculture extension agencies have thoroughly debunked the old garden myth that adding rocks to the bottom of a pot will increase drainage.
Learn more about hard water click here:
brainly.com/question/28178305
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Answer:
elongation of the brass rod is 0.01956 mm
Explanation:
given data
length = 5 cm = 50 mm
diameter = 4.50 mm
Young's modulus = 98.0 GPa
load = 610 N
to find out
what will be the elongation of the brass rod in mm
solution
we know here change in length formula that is express as
δ =
................1
here δ is change in length and P is applied load and A id cross section area and E is Young's modulus and L is length
so all value in equation 1
δ =
δ =
δ = 0.01956 mm
so elongation of the brass rod is 0.01956 mm
Answer:
hello your question is incomplete attached below is the missing equation related to the question
answer : 40.389° , 38.987° , 38° , 39.869° , 40.265°
Explanation:
<u>Determine the friction angle at each depth</u>
attached below is the detailed solution
To calculate the vertical stress = depth * unit weight of sand
also inverse of Tan = Tan^-1
also qc is in Mpa while σ0 is in kPa
Friction angle at each depth
2 meters = 40.389°
3.5 meters = 38.987°
5 meters = 38.022°
6.5 meters = 39.869°
8 meters = 40.265°
Answer:
0, 1, 2, 3, 6, 11, 20, 37, 68
Explanation:
Each number in the series, starting with 3, is the total of the three numbers before it. The following number is the total of the preceding three numbers, according to the pattern. The pattern's next number is 125.