Find the complete solution in the given attachments
Beagle = 494,080
Brash = 295,229.7
Earle = 53,093.4
Heroina = 455,416
Platter = 100,809
38.60% × 1,280,000
44.13% × 669,000
24.81% × 214,000
43.79% × 1,040,000
48.7% × 207,000
hope this helps
a. The percent of change in perimeter is 100% increase
b. The percent of change in Area in 300% increase.
The diagram was missing we have attached for your reference.
Length of the sandbox = 10 ft
Width of the sandbox = 6 ft
We need to find the change in percentage of perimeter and area when the dimensions are doubled.
First we will find the Perimeter and area of the sand box with current dimension.
Now we know that;
Perimeter of rectangle is twice the sum of length and width.
Perimeter of rectangle =
Area of rectangle is length times width.
Perimeter of rectangle =
Now When the dimension are doubled.
Length of the sand box = 20 ft
width of the sandbox = 12 ft
Perimeter of sandbox when dimensions are doubled =
Area of the sand box when dimension are doubled =
Now we need to find the Percent of change in Perimeter and Area.
Percent of change can be calculated by new value minus original value divided by original value multiplied by 100.
Percent of change in Perimeter =
Percent of change in area =
Hence When the dimension are doubled then the percent of change in perimeter is 100% increase and the percent of change in Area in 300% increase.
38.6/100 × 1280000
44.13/100 × 669000
24.81/100 × 214000
43.79/100 × 1040000
48.8/100 × 207000
%Reduction in area = 73.41%
%Reduction in elongation = 42.20%
Original diameter = 12.8 mm
Gauge length = 50.80mm
Diameter at the point of fracture = 6.60 mm (0.260 in.)
Fractured gauge length = 72.14 mm.
%Reduction in Area is given as:
((do/2)² - (d1/2)²)/(do/2)²
Calculating percent reduction in area
do = 12.8mm, d1 = 6.6mm
%RA = ((12.8/2)² - 6.6/2)²)/(12.8/2)²
%RA = 0.734130859375
%RA = 73.41%
Calculating percent reduction in elongation
%Reduction in elongation is given as:
((do) - (d1))/(d1)
do = 72.14mm, d1 = 50.80mm
%RA = ((72.24) - (50.80))/(50.80)
%RA = 0.422047244094488
%RA = 42.20%
% reduction in area==PR=0.734=73.4%
given do=12.8 mm
% reduction in area=((*(do/2)^2)-(*(df/2)^2)))/*(do/2)^2
% reduction in area=73.4%
% elongation=((72.4-50. 8)/50.8)*100=42%
B. temperature, acidity, mineral composition and surface area
Independent variables are those which can be changed and manipulated in the experiment the effect of which can be observed on the changes that occur in the dependent variable. For example the effect of quantity of manure and water on the height of the plant. Here, quantity of manure and water are the independent variables and the height is the dependent variable.
The temperature is the independent variable because the level of temperature can be increased or lowered down manually.
The acidity is the independent variable because the acidity can be manipulated by addition of chemicals that exhibit more hydrogen ions.
Mineral composition can be manipulated and it is an independent variable by treating a mineral with a suitable chemical this will bring a change in the chemical composition of the mineral.
Surface area is the independent variable because it is the total area required for the conduction of the experiment . The surface area can be increased or decreased.
The answers are as follow:
a) 10 mm
b) 12.730 N/
c) 127.307 N/
d1 = 10mm , L1 = 40 mm, L2 = 50 mm, reduction in area = 90% = 0.9
Force = F =1000 N
let us find initial area first, A1 = pi* = 78.55
using reduction in area formula : 0.9 = (A1 - A2 ) / A1
solving it will give, A2 = 0.1 A1 = 7.855
a) The specimen elongation is final length - initial length
50 - 40 = 10 mm
b) Engineering stress uses the original area for all stress calculations,
Engineering stress = force / original area = F / A1 = 1000 / 78.55
Engineering stress = 12.730
c) True stress uses instantaneous area during stress calculations,
True fracture stress = force / final area = F / A2 = 1000 / 7.855
True Fracture stress = 127.30
e) strain = change in length / original length
strain = 10 / 40 = 0.25