Answer:
28.6 kg
Explanation:
The final weight can be calculated from the mixing data and formulae which is given as follows:
cement content =
Computing the parameters and checking the tables gives 28.6 kg.
What is definition ofcomputational fluid Dynamics unstructured grid domaingeometry
1 answer:
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Answer:
the only one that meets the requirements is option C .
Explanation:
The tolerance of a quantity is the maximum limit of variation allowed for that quantity.
To find it we must have the value of the magnitude, its closest value is the average value, this value can be given or if it is not known it is calculated with the formula
x_average = ∑ / n
The tolerance or error is the current value over the mean value per 100
Δx₁ = x₁ / x_average
tolerance = | 100 -Δx₁ 100 |
bars indicate absolute value
let's look for these values for each case
a)
x_average = (2.1700000+ 2.258571429) / 2
x_average = 2.2142857145
fluctuation for x₁
Δx₁ = 2.17000 / 2.2142857145
Tolerance = 100 - 97.999999991
Tolerance = 2.000000001%
fluctuation x₂
Δx₂ = 2.258571429 / 2.2142857145
Δx2 = 1.02
tolerance = 100 - 102.000000009
tolerance 2.000000001%
b)
x_average = (2.2 + 2.29) / 2
x_average = 2,245
fluctuation x₁
Δx₁ = 2.2 / 2.245
Δx₁ = 0.9799554
tolerance = 100 - 97,999
Tolerance = 2.00446%
fluctuation x₂
Δx₂ = 2.29 / 2.245
Δx₂ = 1.0200445
Tolerance = 2.00445%
c)
x_average = (2.211445 +2.3) / 2
x_average = 2.2557225
Δx₁ = 2.211445 / 2.2557225 = 0.9803710
tolerance = 100 - 98.0371
tolerance = 1.96%
Δx₂ = 2.3 / 2.2557225 = 1.024624
tolerance = 100 -101.962896
tolerance = 1.96%
d)
x_average = (2.20144927 + 2.29130435) / 2
x_average = 2.24637681
Δx₁ = 2.20144927 / 2.24637681 = 0.98000043
tolerance = 100 - 98.000043
tolerance = 2.000002%
Δx₂ = 2.29130435 / 2.24637681 = 1.0200000017
tolerance = 2.0000002%
e)
x_average = (2 +2,3) / 2
x_average = 2.15
Δx₁ = 2 / 2.15 = 0.93023
tolerance = 100 -93.023
tolerance = 6.98%
Δx₂ = 2.3 / 2.15 = 1.0698
tolerance = 6.97%
Let's analyze these results, the result E is clearly not in the requested tolerance range, the other values may be within the desired tolerance range depending on the required precision, for the high precision of this exercise the only one that meets the requirements is option C .
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.
