Answer:
T = 5416.67 N
T = -2083.5 N
T = 0
Explanation:
Forward thrust has positive values and reverse thrust has negative values.
part a
Flight speed u = ( 150 km / h ) / 3.6 = 41.67 km / s
The thrust force represents the horizontal or x-component of momentum equation:
Answer: The thrust force T = 5416.67 N
part b
Now the exhaust velocity is now vertical due to reverse thrust application, then it has a zero horizontal component, thus thrust equation is:
Answer: The thrust force T = -2083.5 N reverse direction
part c
Now the exhaust velocity and flight velocity is zero, then it has a zero horizontal component, thus thrust is also zero as there is no difference in two velocities in x direction.
Answer: T = 0 N
Answer:
<em> - 14.943 W/m^2K ( negative sign indicates cooling ) </em>
Explanation:
Given data:
Area of FPC = 4 m^2
temp of water = 60°C
flow rate = 0.06 l/s
ambient temperature = 8°C
exit temperature = 49°C
<u>Calculate the overall heat loss coefficient </u>
Note : heat lost by water = heat loss through convection
m*Cp*dT = h*A * ( T - To )
∴ dT / T - To = h*A / m*Cp ( integrate the relation )
In ( 
) = h* 4 / ( 0.06 * 10^-3 * 1000 * 4180 )
In ( 41 / 52 ) = 0.0159*h
hence h = - 0.2376 / 0.0159
= - 14.943 W/m^2K ( heat loss coefficient )
Answer:
Explanation:
You can utilize barbed clusters to store inadequate grids. On the off chance that there are a great many lines yet each line has just 4 or 5 associations with different segments, at that point as opposed to utilizing a 1000x1000 cluster you can utilize a 1000 line rough exhibit while you simply store the components that the present section has association with another segment. Other utilization can be done on account of query tables. Query tables will be tables which have different qualities concerning a solitary key where the quantity of qualities isn't fixed. Aside from this, barbed clusters have an exceptionally set number of utilization cases. Multidimensional exhibits then again have plenty of utilizations. It is utilized to store a great deal of information reliably on the grounds that the greater part of the information is put away is steady concerning which section compares to what information. Aside from that it very well may be utilized to make thick diagrams or sparse(not effective), plotting information. Another utilization case would be used as an impermanent stockpiling for the figurings that need to tail them and utilize the past information like in powerful programming.
Answer:
$7,778.35
Explanation:
At year 3, the final payment of the remaining balance is equal to the present worth P of the last three payments.
First, calculate the uniform payments A:
A = 12000(A/P, 4%, 5)
= 12000(0.2246) = 2695.2 (from the calculator)
Then take the last three payments as its own cash flow.
To calculate the new P:
P = 2695.2 + 2695.2(P/A, 4%, 2) = 2695.2 + 2695.2(1.886) = 7778.35
Therefore, the final payment is $7,778.35
Answer:
Blank wall
Explanation:
A wall that cannot be moved because it is carrying the weight of the roof is considered a blank wall.
