Answer:1.69*10^12 J
Step-by-step explanation:
From figure above, using triangle ratio
485/755.5=y/l. Cross multiplying 485l=755.5y Divide via 485) hence l= 755.5y/485
Consider a slice volume Vslice= (755.5y/485)^2∆y; recall density =150lb/ft^3
Force slice = 150*755.5^2.y^2.∆y/485^2
From figure 2 in the attachment work done for elementary sclice
Wslice= 150.755.5^2.y^2.∆y.(485-y)/485^2
= (150*755.5^2*y^2)(485-y)∆y/485
To calculate the total work we integrate from y=0 to y= 485
Ie W=[ integral of 150*755.5^2 *y^2(485-y)dy/485] at y=0 and y= 485
Integrating the above
W= 150*755.5^2/485[485*y^3/3-y^4/4] at y= 0 and y=485
W= 150*755.5^2/485(485*485^3/3-484^4/4)-(485.0^3/3-0^4/4)
Work done 1.69*10^12joules
sorry I'm in 5 grade idc what that is tho
1. BC/CD = AC/CE 1. Given
2. <BCA is congruent to <ECD 2. Vertical angles are congruent
3. Tr.ACB is congr to Tr.ECD 3. SAS Similarity
Answer:
Step-by-step explanation:
The base of a cone is simply a circle.
The formula for the area of a circle is A = πr^2.
Example: What is the area of the base of a cone, if the radius of the base is 5 cm?
Answer: A = π(5 cm)^2 = 25π cm^2
Y = mx + b
in this case m = 2 and b = 1 so
equation
y =2x + 1
hope this helps