<span>Nuclear energy can be used to power all of
the above choices. Nuclear power plants produce radioactive waste that must be
stored properly. It is very impossible for a nuclear power plant to have no
waste at all since lots of chemicals are used to create the process as it gives
energy to other machines, weapons such as bombs and powering submarines. Radioactive waste can not be released into
local water supplies since the wastes are very radioactive and may cause
mutation to the fishes and bioaccumulation which will affect humans as well. It
will also cause air pollution if the chemicals are not stored properly.</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:
As the height increases the pressure must increase.
Explanation:
When we add masses to the fluid, the amount of fluid in the tank increases, therefore its height increases and the pressure is described by the expression
P = ρ g h
where rho is constant for a given fluid and h is the height measured from the surface of the fluid.
As the height increases the pressure must increase.
Parallel circuits allow current to flow even if some paths are cut, because a parallel circuit has more than one current-carrying path through it.
Answer:
the mass of water is 0.3 Kg
Explanation:
since the container is well-insulated, the heat released by the copper is absorbed by the water , therefore:
Q water + Q copper = Q surroundings =0 (insulated)
Q water = - Q copper
since Q = m * c * ( T eq - Ti ) , where m = mass, c = specific heat, T eq = equilibrium temperature and Ti = initial temperature
and denoting w as water and co as copper :
m w * c w * (T eq - Tiw) = - m co * c co * (T eq - Ti co) = m co * c co * (T co - Ti eq)
m w = m co * c co * (T co - Ti eq) / [ c w * (T eq - Tiw) ]
We take the specific heat of water as c= 1 cal/g °C = 4.186 J/g °C . Also the specific heat of copper can be found in tables → at 25°C c co = 0.385 J/g°C
if we assume that both specific heats do not change during the process (or the change is insignificant)
m w = m co * c co * (T eq - Ti co) / [ c w * (T eq - Tiw) ]
m w= 1.80 kg * 0.385 J/g°C ( 150°C - 70°C) /( 4.186 J/g°C ( 70°C- 27°C))
m w= 0.3 kg
