Answer:
the correct answer is 36cm2
Answer:
M/N=5/8 or 62.5%
Step-by-step explanation:
when turning the word problem to equation...
![\frac{x}{y} \frac{80}{100}*M=\frac{50}{100}N\\\\ \frac{80M}{100}=\frac{50N}{100}\\ then you should criss-cross\\\100(80M)=100(50N) then you divide both sides by 100)\\80M=50N\\\\\frac{80M}{80}=\frac{50N}{80}\\ M=\frac{50}{80}=\frac{5N}{8}](https://tex.z-dn.net/?f=%5Cfrac%7Bx%7D%7By%7D%20%5Cfrac%7B80%7D%7B100%7D%2AM%3D%5Cfrac%7B50%7D%7B100%7DN%5C%5C%5C%5C%20%5Cfrac%7B80M%7D%7B100%7D%3D%5Cfrac%7B50N%7D%7B100%7D%5C%5C%20%20then%20you%20should%20criss-cross%5C%5C%5C100%2880M%29%3D100%2850N%29%20then%20you%20divide%20both%20sides%20by%20100%29%5C%5C80M%3D50N%5C%5C%5C%5C%5Cfrac%7B80M%7D%7B80%7D%3D%5Cfrac%7B50N%7D%7B80%7D%5C%5C%20%20M%3D%5Cfrac%7B50%7D%7B80%7D%3D%5Cfrac%7B5N%7D%7B8%7D)
since you have the value of M in terms of N you should substitute M by 5N/8.
if you want it as a percent just multiply 5/8 by 100%.
![BY PERCENT=\frac{5}{8}*100%=\frac{500}{8}=62.5%](https://tex.z-dn.net/?f=BY%20PERCENT%3D%5Cfrac%7B5%7D%7B8%7D%2A100%25%3D%5Cfrac%7B500%7D%7B8%7D%3D62.5%25)
=62.5%
Answer:
Percentage Calculator: 15 is what percent of 55? = 27.27.
Step-by-step explanation:
I hope this helps mark me brainlist:)
Answer:
a. 16 slug b. 3.2 ft
Step-by-step explanation:
a. Total mass of the rod
Since the linear density at a point of the rod,λ varies directly as the third power of the measure of the distance of the point form the end, x
So, λ ∝ x³
λ = kx³
Since the linear density λ = 2 slug/ft at then center when x = L/2 where L is the length of the rod,
k = λ/x³ = λ/(L/2)³ = 8λ/L³
substituting the values of the variables into the equation, we have
k = 8λ/L³
k = 8 × 2/4³
k = 16/64
k = 1/4
So, λ = kx³ = x³/4
The mass of a small length element of the rod dx is dm = λdx
So, to find the total mass of the rod M = ∫dm = ∫λdx we integrate from x = 0 to x = L = 4 ft
M = ∫₀⁴dm
= ∫₀⁴λdx
= ∫₀⁴(x³/4)dx
= (1/4)∫₀⁴x³dx
= (1/4)[x⁴/4]₀⁴
= (1/16)[4⁴ - 0⁴]
= (256 - 0)/16
= 256/16
= 16 slug
b. The center of mass of the rod
Let x be the distance of the small mass element dm = λdx from the end of the rod. The moment of this mass element about the end of the rod is xdm = λxdx = (x³/4)xdx = (x⁴/4)dx.
We integrate this through the length of the rod. That is from x = 0 to x = L = 4 ft
The center of mass of the rod x' = ∫₀⁴(x⁴/4)dx/M where M = mass of rod
= (1/4)∫₀⁴x⁴dx/M
= (1/4)[x⁵/5]₀⁴/M
= (1/20)[x⁵]₀⁴/M
= (1/20)[4⁵ - 0⁵]/M
= (1/20)[1024 - 0]/M
= (1/20)[1024]/M
Since M = 16, we have
x' = (1/20)[1024]/16
x' = 64/20
x' = 3.2 ft