Can someone pls help me with this !
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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
Given:
The equation goes through the coordinates (6,5) and (-2,-5).
To find:
The values of A and B.
Solution:
We have,
It goes through the coordinates (6,5) and (-2,-5). It means these points will satisfy the equation.
...(i)
...(ii)
Adding (i) and (ii), we get
Divide both sides by 4.
Put A=5 in (i), we get
Divide both sides by 5.
Therefore, the values of A and B are 5 and -4 respectively.