Answer:
W = -0.480 J
Explanation:
given,
q₁ = 4 μC
q₂ = -4.10 μC
b = 0.381
k = 8.99 × 10⁹ Nm²/C²
![W = [-147.436\times (5.88-2.62)\times 10^{-3}]J](https://tex.z-dn.net/?f=W%20%3D%20%5B-147.436%5Ctimes%20%285.88-2.62%29%5Ctimes%2010%5E%7B-3%7D%5DJ)
W = -0.480 J
Work done by the electric force W = -0.480 J
solution:
radius of steel ball(r)=5cm=0.05m
density of ball =8000kgm
terminal velocity(v)=25m/s^2
density of air( d) =1.29 kgm
now
volume of ball(V)=4/3pir^3=1.33×3.14×0.05^3=0.00052 m^3
density of ball= mass of ball/Volume of ball
or, 8000=m/0.00052
or, m=4.16 kg
weight of the ball (W)= mg=4.16×10=41.6 N
viscous force(F)=6 × pi × eta × r × v
=6×3.14×eta×0.05×25
=23.55×eta
To attain the terminal velocity,
Fiscous force=Weight
or, 23.55× eta = 41.6
or, eta = 1.76
whete eta is the coefficient of viscosity.
Air and water have a good day
Free fall is a special case of motion with constant acceleration, because acceleration due to gravity is always constant and downward. For example, when a ball is thrown up in the air, the ball's velocity is initially upward.
