Refer to the diagram shown below.
The given data is
mass, kg Coordinates. m
------------- -----------------
2 (0, 0)
2 (2, 0)
4 (2, 1)
Total mass, M = 2+2+4 = 8kg
Let (x,y) be the coordinates of M.
Then, taking moments about the origin, we obtain
8x = 2*0 + 2*2 + 4*2 = 12
x = 1.5 m
8y = 2*0 + 2*0 + 4*1 = 4
y = 0.5 m
Answer: (1.5, 0.5) m
Answer:
(a) charge q=5.33 nC
(b) charge density σ=10.62 nC/m²
Explanation:
Given data
radius r=0.20 m
potential V=240 V
coulombs constant k=9×10⁹Nm²/C²
To find
(a) charge q
(b) charge density σ
Solution
For (a) charge q
As

For (b) charge density
As charge density σ is given as:
σ=q/(4πR²)
σ=(5.333×10⁻⁹) / (4π×(0.20)²)
σ=10.62 nC/m²
It would take 57 test tubes to equal a pint