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.
Answer: option c) equal to the angle of reflection.
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Answer:
Explanation:
a )
from lens makers formula

f is focal length , r₁ is radius of curvature of one face and r₂ is radius of curvature of second face
putting the values

1.462 = 2 - 1 / r₂
1 / r₂ = .538
r₂ = 1.86 cm .
= 18.6 mm .
b )
object distance u = 25 cm
focal length of convex lens f = 1.8 cm
image distance v = ?
lens formula



.5555 - .04
= .515
v = 1.94 cm
c )
magnification = v / u
= 1.94 / 25
= .0776
size of image = .0776 x size of object
= .0776 x 10 mm
= .776 mm
It will be a real image and it will be inverted.
Answer:
a) 
b) 
c) 
d) 
Explanation:
<u>Given equation of pressure variation:</u>
![\Delta P= (1.78\ Pa)\ sin\ [(0.888\ m^{-1})x-(500\ s^{-1})t]](https://tex.z-dn.net/?f=%5CDelta%20P%3D%20%281.78%5C%20Pa%29%5C%20sin%5C%20%5B%280.888%5C%20m%5E%7B-1%7D%29x-%28500%5C%20s%5E%7B-1%7D%29t%5D)
We have the standard equation of periodic oscillations:

<em>By comparing, we deduce:</em>
(a)
amplitude:

(b)
angular frequency:


∴Frequency of oscillations:


(c)
wavelength is given by:



(d)
Speed of the wave is gives by:


