Answer:
(a) attached below
(b)

(c) 
(d)
Ω
(e)
and 
Explanation:
Given data:





(a) Draw the power triangle for each load and for the combined load.
°
°
≅ 

≅ 
The negative sign means that the load 2 is providing reactive power rather than consuming
Then the combined load will be


(b) Determine the power factor of the combined load and state whether lagging or leading.

or in the polar form
°

The relationship between Apparent power S and Current I is

Since there is conjugate of current I therefore, the angle will become negative and hence power factor will be lagging.
(c) Determine the magnitude of the line current from the source.
Current of the combined load can be found by


(d) Δ-connected capacitors are now installed in parallel with the combined load. What value of capacitive reactance is needed in each leg of the A to make the source power factor unity?Give your answer in Ω


Ω
(e) Compute the magnitude of the current in each capacitor and the line current from the source.
Current flowing in the capacitor is

Line current flowing from the source is

Answer:
correct option is (A) 0.5
Explanation:
given data
axial column load = 250 kN per meter
footing placed = 0.5 m
cohesion = 25 kPa
internal friction angle = 5°
solution
we know angle of internal friction is 5° that is near to 0°
so it means the soil is almost cohesive soil.
and for a pure cohesive soil
= 0
and we know formula for
is
= (Nq - 1 ) × tan(Ф) ..................1
so here Ф is very less
should be nearest to zero
and its value can be 0.5
so correct option is (A) 0.5
Answer:
The break force that must be applied to hold the plane stationary is 12597.4 N
Explanation:
p₁ = p₂, T₁ = T₂


The heat supplied =
× Heating value of jet fuel
The heat supplied = 0.5 kg/s × 42,700 kJ/kg = 21,350 kJ/s
The heat supplied =
·
= 20 kg/s
The heat supplied = 20*
= 21,350 kJ/s
= 1.15 kJ/kg
T₃ = 21,350/(1.15*20) + 485.03 = 1413.3 K
p₂ = p₁ × p₂/p₁ = 95×9 = 855 kPa
p₃ = p₂ = 855 kPa
T₃ - T₄ = T₂ - T₁ = 485.03 - 280.15 = 204.88 K
T₄ = 1413.3 - 204.88 = 1208.42 K

T₅ = 1208.42*(2/2.333) = 1035.94 K
= √(1.333*287.3*1035.94) = 629.87 m/s
The total thrust =
×
= 20*629.87 = 12597.4 N
Therefore;
The break force that must be applied to hold the plane stationary = 12597.4 N.