Answer:
(a) check attachment
(b)5 m/s²
Explanation:
Given: radius = 2.00mm: density = 2500kg/m³: viscosity of glycerin = 1.5pa: decity of glycerin = 1250kg/m³: g = 10N/kg = 10m/s²: Fdrag = 6πnrv
(a) for answer check attachment.
(b) For the magnitude of the balls initial acceleration:
Initial net force(f) = mg - upthrust
= 
acceleration (a) = 
c.) fromthe force diagram in the attachment; when the ball attains terminal velocity the net force will be zero(0)

d.) For the magnitude of terminal velocity:

e.) when the ball reaches terminal velocity, the acceleration is zero (0)
186 netwons on Mars and 490 netwons on earth
Answer:
Peninsula
Explanation:
Just did the Ed puzzle and got it right
Answer:
Q = 6.33μC
Explanation:
To find the value of the charge Q you take into account both gravitational force and electric force over each ball. By symmetry you can use the fact that both balls experiences the same forces. Hence you only take into account the forces for one ball for the x component and y component:

M: mass of the ball = 0.09kg
T: tension of the string
F_e: electric force between charges
angle = 45°
The electric force is given by:

Q: charge of the balls
r: distance between balls = 2m
You divide both equation in order to eliminate the tension T:

By doing Q the subject of the formula and replacing you obtain:

hence, the charge of the balls is 6.33μC
0.25 m/s squared
hope this helps x