Encrypting your link and protect the link from viruses, malware, thief, etc! Made your link safe to visit.

The snapshots of the program source code and the output is provided below. The source code is provided at the end.

Note:-

1. The main function asks the user to enter the length and width of the plot and passes them as argumnets to the "recommended_litres_fertilizers()" function.

2. The length and width of the plot are passed as arguments to the "recommended_litres_fertilizers()" function. The function calculates the recommended number if litres of fertilizer and returns it to the main function.

Algorithm:-

function recommended_litres_fertilizers(length , width):

area <- length*width;

litres_required <- (area/140);

recommended_litres_required <-(litres_required*1.01400);

return recommended_litres_required;

In case any error is encountered while executing the program , inform in the comments.

1 2 def recommended_litres_fertilizers(length,width): #length and width of the plot are passed as parameters 3 4 total_area =

Output:-

Provide the length of the plot : 500 Provide the width of the plot : 300 Recommended amount of fertilizer to be used on this

Source code:-


def recommended_litres_fertilizers(length,width): #length and width of the plot are passed as parameters

total_area = length*width #area of the plot

fertilizer_required = (total_area/140) # 1 litre of fertilizer if required for 140 metres square of area

recommended_amount = fertilizer_required*1.01400
return recommended_amount

if __name__=="__main__":

length = int(input("Provide the length of the plot : "))
width = int(input("Provide the width of the plot : "))

print("Recommended amount of fertilizer to be used on this plot : {}".format(recommended_litres_fertilizers(length,width)))

ST

Search This Blog

Labels

Report Abuse

QUESTION 6 (a) The bar shown in Figure Q2(a) is subjected to tensile load of 150 Kn. If the stress in the middle portions is limited to 160 N/mm², determine the diameter of the middle portion. Find also the length of the middle portion if the total elongation of the bar is to be 0.25 mm. E E = 2.0 + 105N/mm². (12 marks) 150 KN 10 cm DIA 10 cm DIA 150 KN 45 cm Figure Q6(a) (b) A brass bar, having cross-section area of 900 mm², is subjected to axial forces as shown in Figure Q2(b), in which AB = 0.6 m, BC = 0.8 m, and CD = 1.0 m. Find the total elongation of the bar. E = 1.0 + 105N/mm2 . (8 marks) Page 4 of 5 B D 40 KN 70 KN 20 KN 10 KN Figure Q6(b) (TOTAL = 20 MARKS)

  Question: Show transcribed image text Answer:
SERVER DISCORD FREE CHEGG, COURSEHERO, BARTLEBY https://discord.gg/eGMtZaqSZP https://www.coursehero.com/tutors-problems/Environmental-Science/29241920-Compare-and-contrast-chemical-oxidation-and-chemical-reduction-technol/?justUnlocked=1 ad Oxidation-reduction reaction , also called redox reaction , any chemical reaction in which the oxidation number of a participating chemical species changes. The term covers a large and diverse body of processes. Many oxidation-reduction reactions are as common and familiar as fire, the rusting and dissolution of metals, the browning of fruit, and respiration and photosynthesis—basic life functions.  Oxidation is the  gain  of oxygen. Reduction is the  loss  of oxygen. Most oxidation-reduction (redox) processes involve the transfer of oxygen atoms, hydrogen atoms, or electrons, with all three processes sharing two important characteristics: (1) they are coupled in any oxidation reaction a reciprocal reduction occu...

Contributors