Answer:
Explanation:
The region around a charged particle where another charged particle experiences a force of attraction or repulsion is called electric field.
The strength of electric field is defined as the force experienced by the unit positive test charge.
E = F / q
Electric field strength is a vector quantity and it is measured in newton per coulomb.
Where, F is the force of attraction or repulsion between the two charges and q is the test charge on which the electric field strength is to be calculated.
The strength of electric field is more if the field is strong. It means more be the electric field strength at a point more be the electric field.
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)

The acceleration of this car is equal to 5
.
<u>Given the following data:</u>
- Initial velocity = 0 m/s (assuming it's starting from rest).
To determine the acceleration of this car:
<h3>How to calculate acceleration.</h3>
In Science, the acceleration of an object is calculated by subtracting the initial velocity from its final velocity and dividing by the time.
Mathematically, acceleration is given by this formula:

<u>Where:</u>
- U is the initial velocity.
- is the time measured in seconds.
Substituting the given parameters into the formula, we have;

Acceleration, a = 5 
Read more on acceleration here: brainly.com/question/24728358
Answer:
21.35 cm^3
Explanation:
let the volume at the surface of fresh water is V.
The volume at a depth of 100 m is V' = 2 cm^3
temperature remains constant.
density of water, d = 1000 kg/m^3
Pressure at the surface of fresh water is atmospheric pressure,
P = Po = 1.013 x 10^5 N/m^2
The pressure at depth 100 m is P' = Po + hdg
P' = 
P' = 10.813 x 10^5 N/m^2
Use the Boyle's law
P V = P' V'

V = 21.35 cm^3
Thus, the volume of air bubble at the surface of fresh water is 21.35 cm^3.
A = 1.15m/s2, Vf = 80.0km/h --> we need it in m/s, so:
Vf = 80km/h × 1000m/1km × 1h/3600s
= 22.22m/s
Top speed = Vf, initial speed = Vi
time (t) = V(Vf-Vi) ÷ a
t = (22.22-0)m/s ÷ 1.15m/s2
t = 22.22m/s × s2/1.15m
= 19.32 seconds