Answer:
The three ways thermal energy is transferred are;
1) Conduction
2) Convection
3) Radiation
Explanation:
1) The conduction of the heat from the open flame to the marshmallow is through the direct contact of the flame with the marshmallow, such that the flame the region of the combustion reaction, that produces light and heat touches the marshmallow
2) The convection process is the transfer of heat from the rising heated combustion products, as well as the heated air that rises from the flame
3) The radiation heat transfer is the transfer of the heat from the fire to the marshmallows directly by the heat the moves in the form of electromagnetic waves at temperatures above 1000 K, without the need for a medium, such that the marshmallow can be heated by the heat coming from side of the flame.
Using land according to its capability. protect the soil surface with some form of cover. control runoff before it develops into an erosive force
Answer:
(a) The self inductance, L = 21.95 mH
(b) The energy stored, E = 4.84 J
(c) the time, t = 0.154 s
Explanation:
(a) Self inductance is calculated as;
where;
N is the number of turns = 1000 loops
μ is the permeability of free space = 4π x 10⁻⁷ H/m
l is the length of the inductor, = 45 cm = 0.45 m
A is the area of the inductor (given diameter = 10 cm = 0.1 m)
(b) The energy stored in the inductor when 21 A current ;
(c) time it can be turned off if the induced emf cannot exceed 3.0 V;
Initial volume of mercury is
V = 0.1 cm³
The temperature rise is 35 - 5 = 30 ⁰C = 30 ⁰K.
Because the coefficient of volume expansion is 1.8x10⁻⁴ 1/K, the change in volume of the mercury is
ΔV = (1.8x10⁻⁴ 1/K)*(30 ⁰K)(0.1 cm³) = 5.4x10⁻⁴ cm³
The cross sectional area of the tube is
A = 0.012 mm² = (0.012x10⁻² cm²).
Therefore the rise of mercury in the tube is
h = ΔV/A
= (5.4x10⁻⁴ cm³)/(0.012x10⁻² cm²)
= 4.5 cm
Answer: 4.5 cm
The eaths radius is the correct answer, if you need proof look at nasa's website
