When placed at a certain point, a 4.20 C charge feels an electric force of 2.55 N. What is the magnitude of the electric field a
t that point?
(Remember, magnitudes are always positive.) (Unit = N/C)
(Remember, magnitudes are always positive.) (Unit = N/C)
2 answers:
You might be interested in
Answer:
Tension in the cable is T = 16653.32 N
Explanation:
Give data:
Cross section Area A = 1.3 m^2
Drag coefficient CD = 1.2
Velocity V = 4.3 m/s
Angle made by cable with horizontal =30 degree
Density
Drag force FD is given as
Drag force = 14422.2 N acting opposite to the motion
As cable made angle of 30 degree with horizontal thus horizontal component is take into action to calculate drag force
TCos30 = F_D
T = 16653.32 N