E = <u>kQ</u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>
(r + h)²
where,
k = 9 × 10^9Nm²C^-2
Q = total charge, 300uC = 300 × 10^ -6C
r = 8 × 10^ -2m
h = 16 × 10^ -2m
then,
E = <u>9</u><u>e</u><u>9</u><u> </u><u>*</u><u> </u><u>3</u><u>0</u><u>0</u><u>e</u><u>^</u><u>-</u><u>6</u><u> </u><u> </u><u> </u><u> </u>
(8e^-2 + 16e^-2)²
E = 4687500N/C
Answer:

Explanation:
given,
traffic light weight = 100 N
angle at which the rope is supported = 30°
vertical component of force = ?







Answer:
the angle is given by
Tan theta = 35/59 = 0.59
so theta = Tan ^-1 ( 0.59 )
theta = 30.54 deg.
Explanation:
Th electric force between charges is inversely proportional to the square of distance between them. It means,

Initial distance, r₁ = 2 cm
Final distance, r₂ = 0.25 cm
Initial force, F₁ = 1 N
We need to find the electric force between charges if the new separation of 0.25 cm. So,

So, the new force is 64 N if the separation between charges is 64 N.
Answer:
29.2 ft/s
Explanation:
The distance of the light's projection on the wall
y = 13 tan θ
where θ is the light's angle from perpendicular to the wall.
The light completes one rotation every 3 seconds, that is, 2π in 3 seconds,
Angular speed = w = (2π/3)
w = (θ/t)
θ = wt = (2πt/3)
(dθ/dt) = (2π/3)
y = 13 tan θ
(dy/dt) = 13 sec² θ (dθ/dt)
(dy/dt) = 13 sec² θ (2π/3)
(dy/dt) = (26π/3) sec² θ
when θ = 15°
(dy/dt) = (26π/3) sec² (15°)
(dy/dt) = 29.2 ft/s