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)
It’s solved by using a pretty standard formula for efficiency.
1) <span>yes;2
6*2=12
12*2=24
24*2=48
2)</span><span>Next Term (or nth term) = ar^n-1
</span>
a = first term, i.e. 5
<span>r = common ratio i.e. 3 (as 15/5=3 and 45/15=3 </span>
<span>n = .. </span>
<span>as you already have 1st , 2nd and 3rd terms</span>
<span>substituting now </span>
<span>T4= ar^n-1 </span>
<span>= 5*3^4-1 </span>
<span>= 5*3^3 </span>
<span>= 5*27 </span>
<span>T4 = 135
</span>T5= ar^n-1
<span>= 5*3^5-1 </span>
<span>= 5*3^4 </span>
<span>= 5*81 </span>
<span>T5 = 405 </span>