Answer:
, 
Explanation:
The drag force is equal to:

Where
is the drag coefficient and
is the frontal area, respectively. The work loss due to drag forces is:

The reduction on amount of fuel is associated with the reduction in work loss:

Where
and
are the original and the reduced frontal areas, respectively.

The change is work loss in a year is:
![\Delta W = (0.3)\cdot \left(\frac{1}{2}\right)\cdot (1.20\,\frac{kg}{m^{3}})\cdot (27.778\,\frac{m}{s})^{2}\cdot [(1.85\,m)\cdot (1.75\,m) - (1.50\,m)\cdot (1.75\,m)]\cdot (25\times 10^{6}\,m)](https://tex.z-dn.net/?f=%5CDelta%20W%20%3D%20%280.3%29%5Ccdot%20%5Cleft%28%5Cfrac%7B1%7D%7B2%7D%5Cright%29%5Ccdot%20%281.20%5C%2C%5Cfrac%7Bkg%7D%7Bm%5E%7B3%7D%7D%29%5Ccdot%20%2827.778%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%29%5E%7B2%7D%5Ccdot%20%5B%281.85%5C%2Cm%29%5Ccdot%20%281.75%5C%2Cm%29%20-%20%281.50%5C%2Cm%29%5Ccdot%20%281.75%5C%2Cm%29%5D%5Ccdot%20%2825%5Ctimes%2010%5E%7B6%7D%5C%2Cm%29)


The change in chemical energy from gasoline is:



The changes in gasoline consumption is:





Lastly, the money saved is:


Answer:
Using the above algorithm matches one pair of Ghostbuster and Ghost. On each side of the line formed by the pairing, the number of Ghostbusters and Ghosts are the same, so use the algorithm recursively on each side of the line to find pairings. The worst case is when, after each iteration, one side of the line contains no Ghostbusters or Ghosts. Then, we need n/2 total iterations to find pairings, giving us an P(
)- time algorithm.
Answer:
hello your question is incomplete attached below is the missing equation related to the question
answer : 40.389° , 38.987° , 38° , 39.869° , 40.265°
Explanation:
<u>Determine the friction angle at each depth</u>
attached below is the detailed solution
To calculate the vertical stress = depth * unit weight of sand
also inverse of Tan = Tan^-1
also qc is in Mpa while σ0 is in kPa
Friction angle at each depth
2 meters = 40.389°
3.5 meters = 38.987°
5 meters = 38.022°
6.5 meters = 39.869°
8 meters = 40.265°
On highways, the far left lane is usually the<u> fastest</u> moving traffic.
Answer: Option D.
<u>Explanation:</u>
For the most part, the right lane of a freeway is for entering and leaving the traffic stream. It is an arranging path, for use toward the start and end of your interstate run. The center paths are for through traffic, and the left path is for passing. On the off chance that you are not passing somebody, try not to be driving in the left path.
Regular practice and most law on United States expressways is that the left path is saved for passing and quicker moving traffic, and that traffic utilizing the left path must respect traffic wishing to surpass.
Answer:
(a) The force sustained by the matrix phase is 1802.35 N
(b) The modulus of elasticity of the composite material in the longitudinal direction Ed is 53.7 GPa
(c) The moduli of elasticity for the fiber and matrix phases is 124.8 GPa and 2.2 GPa respectively
Explanation:
Find attachment for explanation