6. Aerial photography is to be taken of a tract of land that is 8 x 8 mi2. Flying height will be 4000 ft above average terrain, and the camera has focal length of 6 inches. If the focal plane opening is 9 x 9 in., and minimum side overlap is 30%, how many flight lines will be needed to cover the tract for the given data

the number of flight lines needed is approximately 72

Step-by-step explanation:

Given the data in the question;

Aerial photography is to be taken of a tract of land that is 8 x 8 mi²

L × B = 8 x 8 mi²

Flying height H = 4000 ft = ( 4000 × 12 )inches = 48000 in

focal length f = 6 in

× b = 9 × 9 in²

side overlap = 30% = 0.3

meaning remaining side overlap = 100% - 30% = 70% = 0.7

{ not end to end overlap }

we take 100% { remaining overlap }

' = 9 × 100% = 9 in

b' = 9 × 70% = 6.3 in

Now the scale will be;

Scale = f/H

we substitute

Scale = 6 in /  48000 in = 1 / 8000

so our scale is; 1 : 8000

⇒ 1 in = 8000 in

⇒ 1 in = (8000 / 63360)mi

⇒ 1 in = 0.126 mi

so since

L × B = 8 x 8 mi²

' = ( 9 × 0.126 mi ) = 1.134 mi

b' = ( 6.3 × 0.126 mi ) = 0.7938 mi

Now we get the flight lines;

N = ( L × B ) / ( ' × b' )

we substitute

N = ( 8 mi × 8 mi ) / ( 1.134 mi × 0.7938 mi )

N = 64 / 0.9001692

N = 71.0977 ≈ 72

Therefore, the number of flight lines needed is approximately 72

725 parts per hour.

step-by-step explanation:

hi there, this answer is actually really simple.

you can set it up as 1450/2 or 3625/5 to find the answer.

they want to know how much per hour is being made, so you just divide.

1450/2=725.

and that's how many are produced per hour.

hope this . : )

the best buy, is the 4- kilogram bag of lollipops for $18.40.

step-by-step explanation:

i found my answer by dividing the cost of each bag by their respective weights.

14.88 divided by 2 = 7.44

17.29 divided by 3 = 5.77

18.4 divided by 4 = 4.6

24.4 divided by 5 = 4.88

by this, i found that the 4 kilo bag is the best buy.