Answer:
14.52 minutes
<u>OR</u>
14 minutes and 31 seconds
Explanation:
Let's first start by mentioning the specific heat of air at constant volume. We consider constant volume and NOT constant pressure because the volume of the room remains constant while pressure may vary.
Specific heat at constant volume at 27°C = 0.718 kJ/kg*K
Initial temperature of room (in kelvin) = 283.15 K
Final temperature (required) of room = 293.15 K
Mass of air in room= volume * density= (4 * 5 * 7) * (1.204 kg/m3) = 168.56kg
Heat required at constant volume: 0.718 * (change in temp) * (mass of air)
Heat required = 0.718 * (293.15 - 283.15) * (168.56) = 1,210.26 kJ
Time taken for temperature rise: heat required / (rate of heat change)
Where rate of heat change = 10000 - 5000 = 5000 kJ/hr
Time taken = 1210.26 / 5000 = 0.24205 hours
Converted to minutes = 0.24205 * 60 = 14.52 minutes
Answer:
Solution:
Note: Refer the diagram
Drag coefficient data for selected objects table at
Hemisphere (open end facing flow), 
Substituting all parameters,
Then,
![\begin{aligned}&V_{b}=V_{w}-\left[\frac{2 F_{R}}{\rho\left(C_{D, w} A_{w}+C_{D, B} A_{b}\right)}\right]^{\frac{1}{2}} \dots\\&V_{w}=24 \times 1000 \times \frac{1}{3600}\\&V_{w}=6.67 \frac{ m }{ s }\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%26V_%7Bb%7D%3DV_%7Bw%7D-%5Cleft%5B%5Cfrac%7B2%20F_%7BR%7D%7D%7B%5Crho%5Cleft%28C_%7BD%2C%20w%7D%20A_%7Bw%7D%2BC_%7BD%2C%20B%7D%20A_%7Bb%7D%5Cright%29%7D%5Cright%5D%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%20%5Cdots%5C%5C%26V_%7Bw%7D%3D24%20%5Ctimes%201000%20%5Ctimes%20%5Cfrac%7B1%7D%7B3600%7D%5C%5C%26V_%7Bw%7D%3D6.67%20%5Cfrac%7B%20m%20%7D%7B%20s%20%7D%5Cend%7Baligned%7D)
And the equation becomes,
![\begin{aligned}&V_{b}=6.67-\left[\frac{2 \times 5.52}{1.23(1.42 \times 1.17+1.2 \times 0.3)}\right]^{\frac{1}{2}}\\&V_{b}=6.67-2.11\\&V_{b}=4.56 \frac{ m }{ s }\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%26V_%7Bb%7D%3D6.67-%5Cleft%5B%5Cfrac%7B2%20%5Ctimes%205.52%7D%7B1.23%281.42%20%5Ctimes%201.17%2B1.2%20%5Ctimes%200.3%29%7D%5Cright%5D%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%5C%5C%26V_%7Bb%7D%3D6.67-2.11%5C%5C%26V_%7Bb%7D%3D4.56%20%5Cfrac%7B%20m%20%7D%7B%20s%20%7D%5Cend%7Baligned%7D)
Thus the floyds travels at 
wind speed.
Explanation:
The square cut seal is what allows the caliper piston to retract back into its housing. Just because you can manually push the piston in, doesn't mean the square cut seal is working. Rapid brake pad wear, brake drag or brake pull are key indicators that the square cut seal is comprised
Answer:
Maximum work under this condition (∆G) = Maximum work under Standard Condition (∆G°) + Activities defining this condition
Explanation:
In this equation, the term DGo provides us with a value for the maximal work we could obtain from the reaction starting with all reactants and products in their standard states, and going to equilibrium. The term DG' provides us with a value for the maximal work we could obtain under the conditions defined by the activities in the logarithmic term. The logarithmic term can be seen as modifying the value under standard conditions to account for the actual conditions. In describing the work available in metabolic processes, we are concerned with the actual conditions in the reaction medium (whether that is a test-tube, or the cell cytoplasm); the important term is therefore DG'. If we measure the actual activities (in practice, we make do with concentrations), and look up a value for DGo in a reference book, we can calculate DG' from the above equation.
Values for DGo provide a useful indication through which we can compare the relative work potential from different processes, because they refer to a standard set of conditions.
Therefore both phrases describe the Biochemical and Chemical Standard State
Answer:
Tech B
Explanation:
Bleeding the master cylinder on the bench does nothing for the air in the brake lines on the vehicle. The bench bleeding is a preferred first step, but bleeding the rest of the brake system is also required. A final check of proper operation on the vehicle should also be accomplished.
Tech B is correct.
