List three types of documents that assist an engineer during reverse engineering and explain their functions.
2 answers:
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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:
vB = - 0.176 m/s (↓-)
Explanation:
Given
(AB) = 0.75 m
(AB)' = 0.2 m/s
vA = 0.6 m/s
θ = 35°
vB = ?
We use the formulas
Sin θ = Sin 35° = (OA)/(AB) ⇒ (OA) = Sin 35°*(AB)
⇒ (OA) = Sin 35°*(0.75 m) = 0.43 m
Cos θ = Cos 35° = (OB)/(AB) ⇒ (OB) = Cos 35°*(AB)
⇒ (OB) = Cos 35°*(0.75 m) = 0.614 m
We apply Pythagoras' theorem as follows
(AB)² = (OA)² + (OB)²
We derive the equation
2*(AB)*(AB)' = 2*(OA)*vA + 2*(OB)*vB
⇒ (AB)*(AB)' = (OA)*vA + (OB)*vB
⇒ vB = ((AB)*(AB)' - (OA)*vA) / (OB)
then we have
⇒ vB = ((0.75 m)*(0.2 m/s) - (0.43 m)*(0.6 m/s) / (0.614 m)
⇒ vB = - 0.176 m/s (↓-)
The pic can show the question.
