Answer:
In Rankine 524.07°R
In kelvin 291 K
In Fahrenheit 64.4°F
Explanation:
We have given temperature 18°C
We have to convert this into Rankine R
From Celsius to Rankine we know that 
We have to convert 18°C
So 
Conversion from Celsius to kelvin
We have to convert 18°C

Conversion of Celsius to Fahrenheit
Explanation:
1) Work done = force x distance x cos(θ)
= 0.15 x 6 x cos(30)
= 0.779
2) Ek = ½mv²
v = acceleration due to gravity so 9.81
Ek = ½(2)(9.81)²
Ek = 96.2361
3) v = (√(2em)) / m
= (√(2(96.2361)(2)) / 2
= 9.81 so especially with no time given, I can only assume the acceleration due to gravity but take it with a pinch of salt.
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.
Answer:
D) 1.04 Btu/s from the liquid to the surroundings.
Explanation:
Given that:
flow rate (m) = 2 lb/s
liquid specific enthalpy at the inlet (
Btu/lb)
liquid specific enthalpy at the exit (
Btu/lb)
initial elevation (
)
final elevation (
)
acceleration due to gravity (g) = 32.174 ft/s²
= 3 Btu/s
The energy balance equation is given as:
![Q_{cv}-W{cv}+m[(h_1-h_2)+(\frac{V_1^2-V_2^2}{2})+g(z_1-z_2)]=0](https://tex.z-dn.net/?f=Q_%7Bcv%7D-W%7Bcv%7D%2Bm%5B%28h_1-h_2%29%2B%28%5Cfrac%7BV_1%5E2-V_2%5E2%7D%7B2%7D%29%2Bg%28z_1-z_2%29%5D%3D0)
Since kinetic energy effects are negligible, the equation becomes:
![Q_{cv}-W{cv}+m[(h_1-h_2)+g(z_1-z_2)]=0](https://tex.z-dn.net/?f=Q_%7Bcv%7D-W%7Bcv%7D%2Bm%5B%28h_1-h_2%29%2Bg%28z_1-z_2%29%5D%3D0)
Substituting values:
![Q_{cv}-(-3)+2[(40.09-40.94)+\frac{32.174(0-100)}{778*32.174} ]=0\\Q_{cv}+3+2[-0.85-0.1285 ]=0\\Q_{cv}+3+2(-0.9785)=0\\Q_{cv}+3-1.957=0\\Q_{cv}+1.04=0\\Q_{cv}=-1.04\\](https://tex.z-dn.net/?f=Q_%7Bcv%7D-%28-3%29%2B2%5B%2840.09-40.94%29%2B%5Cfrac%7B32.174%280-100%29%7D%7B778%2A32.174%7D%20%5D%3D0%5C%5CQ_%7Bcv%7D%2B3%2B2%5B-0.85-0.1285%20%5D%3D0%5C%5CQ_%7Bcv%7D%2B3%2B2%28-0.9785%29%3D0%5C%5CQ_%7Bcv%7D%2B3-1.957%3D0%5C%5CQ_%7Bcv%7D%2B1.04%3D0%5C%5CQ_%7Bcv%7D%3D-1.04%5C%5C)
The heat transfer rate is 1.04 Btu/s from the liquid to the surroundings.