The phrase "positive to positive, negative to ground" is correct when jump starting a car.
1 answer:
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Answer:
# kelvin_to_celsius function is defined
# it has value_kelvin as argument
def kelvin_to_celsius(value_kelvin):
# value_celsius is initialized to 0.0
value_celsius = 0.0
# value_celsius is calculated by
# subtracting 273.15 from value_kelvin
value_celsius = value_kelvin - 273.15
# value_celsius is returned
return value_celsius
# celsius_to_kelvin function is defined
# it has value_celsius as argument
def celsius_to_kelvin(value_celsius):
# value_kelvin is initialized to 0.0
value_kelvin = 0.0
# value_kelvin is calculated by
# adding 273.15 to value_celsius
value_kelvin = value_celsius + 273.15
# value_kelvin is returned
return value_kelvin
value_c = 0.0
value_k = 0.0
value_c = 10.0
# value_c = 10.0 is used to test the function celsius_to_kelvin
# the result is displayed
print(value_c, 'C is', celsius_to_kelvin(value_c), 'K')
value_k = 283.15
# value_k = 283.15 is used to test the function kelvin_to_celsius
# the result is displayed
print(value_k, 'is', kelvin_to_celsius(value_k), 'C')
Explanation:
Image of celsius_to_kelvin function used as guideline is attached
Image of program output is attached.
Answer:
attached below
Explanation:
a) G(s) = 1 / s( s+2)(s + 4 )
Bode asymptotic magnitude and asymptotic phase plots
attached below
b) G(s) = (s+5)/(s+2)(s+4)
phase angles = tan^-1 w/s , -tan^-1 w/s , tan^-1 w/4
attached below
c) G(s)= (s+3)(s+5)/s(s+2)(s+4)
solution attached below
Answer:
Explanation:
Given that:
<u>At state 1:</u>
Pressure P₁ = 20 bar
Volume V₁ = 0.03
From the tables at saturated vapour;
Temperature T₁ = 212.4⁰ C ; = 0.0996
/ kg
The mass inside the cylinder is m = 0.3 kg, which is constant.
The specific internal energy u₁ = ug₁ = 2599.2 kJ/kg
<u>At state 2:</u>
Temperature T₂ = 200⁰ C
Since the 1 - 2 occurs in an isochoric process v₂ = v₁ = 0.099 / kg
From temperature T₂ = 200⁰ C
Since , the saturated pressure at state 2 i.e. P₂ = 15.5 bar
Mixture quality
At temperature T₂, the specific internal energy , also
Thus,
<u>At state 3:</u>
Temperature
Specific volume
Thus; ,
SInce , therefore, the phase is in a superheated vapour state.
From the tables of superheated vapour tables; at and T₃ = 200⁰ C
The pressure = 10 bar and v =0.206
The specific internal energy at the pressure of 10 bar = 2622.3 kJ/kg
The changes in the specific internal energy is:
= (2210.686 - 2599.2) kJ/kg
= -388.514 kJ/kg
≅ - 389 kJ/kg
= (2622.3 - 2210.686) kJ/kg
= 411.614 kJ/kg
≅ 410 kJ/kg
We can see the correct sketches of the T-v plot showing the diagrammatic expression in the image attached below.
