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
Answer:
6 Ω
Explanation:
Two resistors in parallel = 3.75 ohm when one is 10 ...
r1 * r2 / (r1+r2) = 3.75
10 r2 / ( 10 + r2 )= 3.75
10 r2 = 37.5 + 3.75r2
6.25 r2 = 37.5
r2 = 6Ω
Answer:
The distance the plane covered while it was accelerating is 80,633.3 m
Explanation:
Given;
initial velocity of the plane, u = 90 m/s
acceleration of the plane, a = 1.5 m/s²
final velocity of the plane, v = 500 m/s
The distance covered by the plane is given as;
v² = u² + 2ad
where;
d is the distance covered by the plane;
500² = 90² + 2(1.5)d
500² - 90² = 3d
241900 = 3d
d = 241900 / 3
d = 80,633.3 m
Therefore, the distance the plane covered while it was accelerating is 80,633.3 m
Momentum = mass x acceleration
5 x 2 = 10 kg. m/s
