Answer:
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Explanation:
Answer:
Area of Circle = 78.5398
Surface Area of Sphere = 1.2566 x 10^3 = 1256.6 ft
Volume of Sphere = 33.5103 ft
Explanation:
Please find below the written MatLab script used to solve the problem. I had to define r in each case to solve for the Area of the circle, the surface area and the volume of the Sphere.
r=5; % define r as 5
a=pi*r^2;% calculate the area of the circle
AreaOfCircle=a
r=10; % define r and 10 ft
sa=4*pi*r^2; %Calculate the surface area of the sphere
SphereSurfaceArea=sa
r=2;% define r as 2 ft
vs=(4/3)*pi*r^3;% Calculate the volume of the sphere
VolumeShere=vs
Answer:
Given that
LHS of above given equation have dimension ![[M^oL^{1}T^{-1}]](https://tex.z-dn.net/?f=%5BM%5EoL%5E%7B1%7DT%5E%7B-1%7D%5D)
.
Now find the dimension of RHS
Dimension of P = ![[ML^{-1}T^{-2}]](https://tex.z-dn.net/?f=%5BML%5E%7B-1%7DT%5E%7B-2%7D%5D)
.
Dimension of d= ![[M^{0}L^{1}T^{0}]](https://tex.z-dn.net/?f=%5BM%5E%7B0%7DL%5E%7B1%7DT%5E%7B0%7D%5D)
.
Dimension of μ = ![[ML^{-1}T^{-1}]](https://tex.z-dn.net/?f=%5BML%5E%7B-1%7DT%5E%7B-1%7D%5D)
.
Dimension of L= .
So
![\dfrac{\Delta Pd^2}{32\mu L}=\dfrac{[ML^{-1}T^{-2}].[M^{0}L^{1}T^{0}]^2}{[ML^{-1}T^{-1}].[M^{0}L^{1}T^{0}]}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5CDelta%20Pd%5E2%7D%7B32%5Cmu%20L%7D%3D%5Cdfrac%7B%5BML%5E%7B-1%7DT%5E%7B-2%7D%5D.%5BM%5E%7B0%7DL%5E%7B1%7DT%5E%7B0%7D%5D%5E2%7D%7B%5BML%5E%7B-1%7DT%5E%7B-1%7D%5D.%5BM%5E%7B0%7DL%5E%7B1%7DT%5E%7B0%7D%5D%7D)
![\dfrac{\Delta Pd^2}{32\mu L}=[M^0L^{1}T^{-1}]](https://tex.z-dn.net/?f=%5Cdfrac%7B%5CDelta%20Pd%5E2%7D%7B32%5Cmu%20L%7D%3D%5BM%5E0L%5E%7B1%7DT%5E%7B-1%7D%5D)
It means that both sides have same dimensions.
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