The electric potential energy of the charge is reduced because it decreases with increase in the distance between charges.
<h3>What is electric potential energy?</h3>
Electric potential energy can be defined as the energy needed to move a charge against an electric field.
It is calculated using the formula;
U = Kq1 q2 ÷ r
Where Q = electric potential energy
k = Coulombs constant
q1 and q2 = charges
r = distance of separation
Electric potential energy is inversely proportional to the distance of separation of the charges.
If the distance of the charges changes from 3mm to 6mm, then the electric potential energy of the charges is reduced because it decreases with increase in the distance of the charges.
Therefore, the electric potential energy of the charge is reduced because it decreases with increase in the distance between charges.
Learn more about electric potential energy here:
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<span>In a 2-dimensional coordinate system, the x- and y-axes
are typically perpendicular to each other. (C) </span>
Answer:
(a) the electrical power generated for still summer day is 1013.032 W
(b)the electrical power generated for a breezy winter day is 1270.763 W
Explanation:
Given;
Area of panel = 2 m × 4 m, = 8m²
solar flux GS = 700 W/m²
absorptivity of the panel, αS = 0.83
efficiency of conversion, η = P/αSGSA = 0.553 − 0.001 K⁻¹ Tp
panel emissivity , ε = 0.90
Apply energy balance equation to determine he electrical power generated;
transferred energy + generated energy = 0
(radiation + convection) + generated energy = 0
![[\alpha_sG_s-\epsilon \alpha(T_p^4-T_s^4)]-h(T_p-T_\infty) - \eta \alpha_s G_s = 0](https://tex.z-dn.net/?f=%5B%5Calpha_sG_s-%5Cepsilon%20%5Calpha%28T_p%5E4-T_s%5E4%29%5D-h%28T_p-T_%5Cinfty%29%20-%20%5Ceta%20%5Calpha_s%20G_s%20%3D%200)
![[\alpha_sG_s-\epsilon \alpha(T_p^4-T_s^4)]-h(T_p-T_\infty) - (0.553-0.001T_p)\alpha_s G_s](https://tex.z-dn.net/?f=%5B%5Calpha_sG_s-%5Cepsilon%20%5Calpha%28T_p%5E4-T_s%5E4%29%5D-h%28T_p-T_%5Cinfty%29%20-%20%280.553-0.001T_p%29%5Calpha_s%20G_s)
(a) the electrical power generated for still summer day

![[0.83*700-0.9*5.67*10^{-8}(T_p_1^4-308^4)]-10(T_p_1-308) - (0.553-0.001T_p_1)0.83*700 = 0\\\\3798.94-5.103*10^{-8}T_p_1^4 - 9.419T_p_1 = 0\\\\Apply \ \ iteration \ method \ to \ solve \ for \ T_p_1\\\\T_p_1 = 335.05 \ k](https://tex.z-dn.net/?f=%5B0.83%2A700-0.9%2A5.67%2A10%5E%7B-8%7D%28T_p_1%5E4-308%5E4%29%5D-10%28T_p_1-308%29%20-%20%280.553-0.001T_p_1%290.83%2A700%20%3D%200%5C%5C%5C%5C3798.94-5.103%2A10%5E%7B-8%7DT_p_1%5E4%20-%209.419T_p_1%20%3D%200%5C%5C%5C%5CApply%20%5C%20%20%5C%20iteration%20%5C%20method%20%5C%20to%20%5C%20solve%20%5C%20for%20%5C%20T_p_1%5C%5C%5C%5CT_p_1%20%3D%20335.05%20%5C%20k)

(b)the electrical power generated for a breezy winter day

![[0.83*700-0.9*5.67*10^{-8}(T_p_2^4-258^4)]-10(T_p_2-258) - (0.553-0.001T_p_2)0.83*700 = 0\\\\8225.81-5.103*10^{-8}T_p_2^4 - 29.419T_p_2 = 0\\\\Apply \ \ iteration \ method \ to \ solve \ for \ T_p_2\\\\T_p_2 = 279.6 \ k](https://tex.z-dn.net/?f=%5B0.83%2A700-0.9%2A5.67%2A10%5E%7B-8%7D%28T_p_2%5E4-258%5E4%29%5D-10%28T_p_2-258%29%20-%20%280.553-0.001T_p_2%290.83%2A700%20%3D%200%5C%5C%5C%5C8225.81-5.103%2A10%5E%7B-8%7DT_p_2%5E4%20-%2029.419T_p_2%20%3D%200%5C%5C%5C%5CApply%20%5C%20%20%5C%20iteration%20%5C%20method%20%5C%20to%20%5C%20solve%20%5C%20for%20%5C%20T_p_2%5C%5C%5C%5CT_p_2%20%3D%20279.6%20%5C%20k)

Answer
given,
mass of copper rod = 1 kg
horizontal rails = 1 m
Current (I) = 50 A
coefficient of static friction = 0.6
magnetic force acting on a current carrying wire is
F = B i L
Rod is not necessarily vertical


the normal reaction N = mg-F y
static friction f = μ_s (mg-F y )
horizontal acceleration is zero


B_w = B sinθ
B_d = B cosθ
iLB cosθ= μ_s (mg- iLB sinθ)





B = 0.1 T