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ra1l [238]
3 years ago
5

T he area of a circle is pr 2. Define r as 5, then find the area of a circle,using MATLAB®.(b) The surface area of a sphere is 4

pr 2. Find the surface area of a spherewith a radius of 10 ft.(c) The volume of a sphere is 4/3pr 3. Find the volume of a sphere with aradius of 2 ft.
Engineering
2 answers:
horsena [70]3 years ago
4 0

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

aksik [14]3 years ago
3 0

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

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Explanation:

Given.

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Thermal diffusivity of brass \alpha = 33.9\times 10^{-6} m^2/s

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Plate remain in the oven t =10 min  

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The thermal properties of the plate are constant.

The heat transfer coefficient is constant and uniform over the entire surface.

The Fourier number is > 0.2 so that the one-term approximate solutions (or the transient  temperature charts) are applicable (this assumption will be verified).

The Biot number for this process Bi = \frac{hL}{k}\\\\Bi=\frac{(80 W/m^2.°C)(0.015 m)}{(110 W/m.°C)}\\=Bi =0.0109

The constants \lambda_1 and A_1 corresponding to this Biot are, from 11-2 tables.

The interpolation method used to find the

\lambda_1=0.1039  and A_1=1.0018
  

The Fourier number \tau=\frac{\alpha t}{L^2}\\\\\tau=\frac{(33.9\times 10^{-6} m^2/s)(10 min \times 60 s/min)}{(0.015m)^2}
\\\\\tau=90.4>0.2

Therefore, the one-term approximate solution (or the transient temperature charts) is applicable.

Then the temperature at the surface of the plates becomes

\theta(L,t)_{wall}=\frac{T(x,t)-T_{\infty}}{(T_i-T_{\infty})}\\\\\theta(L,t)_{wall}=A_1e^{-\lambda_1^2\tau}\cos(\lambda_1L/L)\\\\\theta(L,t)_{wall}=(1.0018)e^{-(0.1039^2(90.4))}\cos(0.1039)\\\\\theta(L,t)_{wall}=0.378\\\\\frac{T(L,t)-700}{25-700}=0.378\\\\T(L,t)=445°C

3 0
3 years ago
Read 2 more answers
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