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.
Ok ok ok ok ok ok ok ok ok
Il write down the answers of the blanks which are as follows:-
Weight
Electrical
Magnetic
Force
Newtons
Mass
Kilograms
Hope this helps!
Answer:
r = 0.05 m = 5 cm
Explanation:
Applying ampere's law to the wire, we get:

where,
r = distance of point P from wire = ?
μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²
I = current = 2 A
B = Magnetic Field = 8 μT = 8 x 10⁻⁶ T
Therefore,

<u>r = 0.05 m = 5 cm</u>
Answer:
given
y=6.0sin(0.020px + 4.0pt)
the general wave equation moving in the positive directionis
y(x,t) = ymsin(kx -?t)
a) the amplitude is
ym = 6.0cm
b)
we have the angular wave number as
k = 2p /?
or
? = 2p / 0.020p
=1.0*102cm
c)
the frequency is
f = ?/2p
= 4p/2p
= 2.0 Hz
d)
the wave speed is
v = f?
= (100cm)(2.0Hz)
= 2.0*102cm/s
e)
since the trignometric function is (kx -?t) , sothe wave propagates in th -x direction
f)
the maximum transverse speed is
umax =2pfym
= 2p(2.0Hz)(6.0cm)
= 75cm/s
g)
we have
y(3.5cm ,0.26s) = 6.0cmsin[0.020p(3.5) +4.0p(0.26)]
= -2.0cm