Answer:
See attachment below
Explanation:
The escape velocity of a rocket from the earth is 11kms^-1. Taking the acceleration due to gravity as 9.8ms^-2, calculate the ra
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Answer:
body position 4 is (-1,133, -1.83)
Explanation:
The concept of center of gravity is of great importance since in this all external forces are considered applied, it is defined by
x_cm = 1 /M ∑ m_{i}
y_cm = 1 /M ∑ y_{i} mi
Where M is the total mass of the body, mi is the mass of each element
give us the mass and position of this masses
body 1
m1 = 2.00 ka
x1 = 0 me
y1 = 0 me
body 2
m2 = 2.20 kg
x2 = 0m
y2 = 5 m
body 3
m3 = 3.4 kg
x3 = 2.00 m
y3 = 0
body 4
m4 = 6 kg
x4=?
y4=?
mass center position
x_cm = 0
y_cm = 0
let's apply to the equations of the initial part
X axis
M = 2.00 + 2.20 + 3.40
M = 7.6 kg
0 = 1 / 7.6 (2 0 + 2.2 0 + 3.4 2 + 6 x4)
x4 = -6.8 / 6
x4 = -1,133 m
Axis y
0 = 1 / 7.6 (2 0 + 2.20 5 +3.4 0 + 6 y4)
y4 = -11/6
y4 = -1.83 m
body position 4 is (-1,133, -1.83)
Answer: See the explanation below.
Explanation: For this assignment, I chose to display how eclipses are created.
My model was made utilizing a 3D displaying device program for all intents and purposes. The items utilized are three models I made for this presentation, Earth, the moon, and the sun. These three models will be utilized for the showcase.
The light that shines from the sun would create a shadow on the moon. The moon would then catch the light that should've arrived on Earth, making the shadow we call an eclipse. Earth gets a shadow of the moon and the remainder of Earth is lit up from the rest of the light, making an eclipse.
The individual I demonstrated my project to was [<em>Someone you know</em>], [<em>Pronoun</em>] said it precisely took after the occasion of an eclipse. The light from the sun being shined on to the moon rather than the Earth, creating the shadow we call an eclipse.
