Answer:
E = 3544.44 N/C
Explanation:
Given:
- charge Q = 2.2 *10^-6 C
- Length L = 1.3 m
Find:
The Electric Field strength E @ a = 1.8 m
Solution:
- The differential electric field dE due to infinitesimal charge dq can be considered as a point charge at a distance of r is given by:
dE = k*dq / r^2
- The charge Q is spread over entire length L, hence:
dq = (Q / L ) * dx
-The resulting dE:
dE = (k*Q/L)*(dx / r^2)
- point P lies on the x- axis with distance (x+a) from differential charge from:
dE = (k*Q/L)*(dx / (x+a)^2)
- Integrate dE over length 0 to L
E = (-k*Q/L)*( 1 / (x+a) )
E = (-k*Q/L)* (1 / a - 1 / (L+a))
E = (-k*Q/L)* (L / a(L+a))
E = (k*Q / a(L+a))
- Evaluate E @ a = 1.8 m
E =(8.99*10^9 * 2.2*10^-6 / 1.8*(1.3+1.8))
E = 3544.44 N/C
Watt is a unit for power that is also equal to J/s. We therefore need to convert the minutes to seconds first before answering. Every minute is comprised of 60 seconds. Therefore, 3 minutes are composed of 180 seconds. Multiplying the number of seconds to the given power will give us,
Work = Power x time
= (1500 J/s) x (180 s)
= 270,000 J
Therefore, the answer is letter D.
<span>Mass of the ball is m = 0.10kg
Initial speed of the Ball v = 15m/s
a. When the ball is at maximum height the velocity is 0
Momentum of ball = mass x velocity
Momentum = 0.10kg x 0 = 0
b. Getting the maximum height,
Using the conservation of energy equation KEinitial = mgh
1/2mVin^2 = mgh => h = v^2/2g
h = 15^2/2x9.8 = 11.48m => Half Height h = 5.96m
Applying the conservation of energy equation at halfway V^2 = 2gh
V = square root of (2x9.8x5.96) => V = square root of (116.816)
So the velocity at the half way V = 10.81 m/s
Momentum M = m x V => M = 0.10 x 10.81 => M = 1.081kg-m/s</span>
