Question 3. Assign boston_under_10 and manila_under_10 to the percentage of rides that are less than 10 minutes in their respect
ive metropolitan areas. Use the height variables provided below in order to compute the percentages. Your solution should only use height variables, numbers, and mathematical operations. You should not access the tables boston and manila in any way.
1 answer:
Answer:
Enter the following code to get the required conditions for the answer.
boston_under_5_height = 1.2
manila_under_5_height = 0.6
boston_5_to_under_10_height = 3.2
manila_5_to_under_10_height = 1.4
boston_under_10 = boston_under_10 = 5*boston_under_5_height + 5*boston_5_to _under_10_height
manila_under_10 = manila_under_10 = 5*manila_under_5_height + 5*manila_5_to _under_10_height
Explanation:
Kindly note that question in complete as it belong to the topic of table manipulation and visualization topic. The question asks about the company called Uber and their data extraction from the website called movements.uber.com where data was extracted for 200,000 weekdays in the respective cities of Manila, Philippines and Boston, Massachusetts. Images attached contains the histograms generated for the rides in manila and boston.
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Answer:
The answer to the question is
The heat transferred in the process is -274.645 kJ
Explanation:
To solve the question, we list out the variables thus
R-134a = Tetrafluoroethane
Intitial Temperaturte t₁ = 100 °C
Initial pressure = 3.5 bar = 350 kPa
For closed system we have m₁ = m₂ = m
ΔU = m×(u₂ - u₁) = ₁Q₂ -₁W₂
For constant pressure process we have
Work done = W = = P×ΔV = P × (V₂ - V₁) = P×m×(v₂ - v₁)
From the tables we have
State 1 we have h₁ = (490.48 +489.52)/2 = 490 kJ/kg
State 2 gives h₂ = 206.75 + 0.75 × 194.57= 352.6775 kJ/kg
Therefore Q₁₂ = m×(u₂ - u₁) + W₁₂ = m × (u₂ - u₁) + P×m×(v₂ - v₁)
= m×(h₂ - h₁) = 2.0 kg × (352.6775 kJ/kg - 490 kJ/kg) =-274.645 kJ
Answer:
In Btu:
Q=0.001390 Btu.
In Joule:
Q=1.467 J
Part B:
Temperature at midpoint=274.866 C
Explanation:
Thermal Conductivity=k=30 (Btu/hr)/(ft ⋅ °F)=
Thermal Conductivity is SI units:
Length=20 cm=0.2 m= (20*0.0328) ft=0.656 ft
Radius=4/2=2 mm =0.002 m=(0.002*3.28)ft=0.00656 ft
T_1=500 C=932 F
T_2=50 C= 122 F
Part A:
In Joules (J)
Heat Q is:
In Btu:
Heat Q is:
PArt B:
At midpoint Length=L/2=0.1 m
On rearranging: