Answer:231.16 N/C
Explanation:
Given
Electric Flux
Area(A)
Given Field point above 
Therefore angle between Area vector Electric Field =90-31.6=
We know that Flux is given by



E=231.16 N/C
Answer:
81 J.
Explanation:
From the question given above, the following data were obtained:
Specific heat capacity (C) = 0.45 J/gK.
Temperature change (ΔT) = 15 K
Mass = 12 g
Heat required (Q) =?
The heat required to raise the temperature of iron can be obtained as illustrated below:
Q = MCΔT
Q = 12 × 0.45 × 15
Q = 81 J
Therefore, the heat required to raise the temperature of the iron is 81 J.
Answer:
F_n = 5.65E-11 N
d = 1.20682E-31 m
Explanation:
F = 3.8E-09 N
where
m = Mass of electron = 9.109E−31 kilograms
G = Gravitational constant = 6.67E-11 m³/kgs²
x = Distance between them

For 

Dividing the above equations we get

F_n = 5.65E-11 N

d = 1.20682E-31 m
(0.5)×(0squared)×(3)=(1.5j)