Answer:
(1) The maximum air temperature is 1383.002 K
(2) The rate of heat addition is 215.5 kW
Explanation:
T₁ = 17 + 273.15 = 290.15
T₂ = 290.15 × 3.17767 = 922.00139
Therefore,
T₃ = T₂×1.5 = 922.00139 × 1.5 = 1383.002 K
The maximum air temperature = T₃ = 1383.002 K
(2)
Therefore;
Q₁ = 1.005(1383.002 - 922.00139) = 463.306 kJ/jg
Heat rejected per kilogram is given by the following relation;
= 0.718×(511.859 - 290.15) = 159.187 kJ/kg
The efficiency is given by the following relation;
Where:
β = Cut off ratio
Plugging in the values, we get;
Therefore;
Heat supplied = 
Therefore, heat supplied = 215491.064 W
Heat supplied ≈ 215.5 kW
The rate of heat addition = 215.5 kW.
Metal ores
Explanation:
in an area where subduction has occurred in times past, metal ores are likely to be found.
Metallic ores find subduction zone regions very favorable to crystallize out of a magma.
- Ores have different modes of formation.
- Typically, they are found in hydrothermal vents and black smokers of igneous intrusives.
- These are igneous terrains where metallic sulfides and other minerals crystallize out of magmatic body.
- Metals in magma usually have large sizes and do not partition easily in the melt.
At a subduction zone, partial melting of the subducting plate forces magma into nearby country rock as an intrusive and to the ocean floor where they form black smokers.
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Answer:
The permittivity of rubber is 
Explanation:
From the question we are told that
The magnitude of the point charge is 
The diameter of the rubber shell is 
The Electric field inside the rubber shell is 
The radius of the rubber is mathematically evaluated as
Generally the electric field for a point is in an insulator(rubber) is mathematically represented as
Where 
is the permittivity of rubber
=> 
=> 
substituting values
Answer:
Explanation:
we know that half life of an element is
T=0.693/λ
where λ is decay constant in order to find decay constant
λ=0.693/T
λ=0.693/8.04
λ=0.086
To solve this problem it is necessary to apply the concepts related to the principle of superposition and the equations of destructive and constructive interference.
Constructive interference can be defined as
Where
m= Any integer which represent the number of repetition of spectrum
= Wavelength
d = Distance between the slits.
= Angle between the difraccion paterns and the source of light
Re-arrange to find the distance between the slits we have,
Therefore the number of lines per millimeter would be given as
Therefore the number of the lines from the grating to the center of the diffraction pattern are 380lines per mm
