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:critical stress= 20.23 MPa
Explanation:
Since there was an internal crack, we will divide the length of the internal crack by 2
Length of internal crack, a = 0.7mm,
Half length = 0.7mm/2= 0.35mm changing to meters becomes
0.35/ 1000= 0.35 x 10 ^-3m
The formulae for critical stress is calculated using
σC = (2Eγs /πa) ¹/₂
σC = critical stress=?
Given
E= Modulus of Elasticity= 225GPa =225 x 10 ^ 9 N/m²
γs= Specific surface energy = 1.0 J/m2 = 1.0 N/m
a= Half Length of crack=0.35 x 10 ^-3m
σC= (2 x 225 x 10 ^ 9 N/m² x 1.0 N/m /π x 0.35 x 10 ^-3m)¹/₂
=(4.5 x 10^11/π x 0.35 x 10 ^-3)¹/₂
=(4.0920 x10 ^14)¹/₂
σC=20.23 x10^6 N/m² = 20.23 MPa
Answer:
They both work together in order to create a good experience for the passenger, but I think that the answer is changing acceleration because when the rollercoaster changes acceleration it gives it speed.
Explanation:
I really hope this helps.
Answer:
Refrigerant R-134a is to be cooled by waterin a condenser.The refrigerant enters thecondenser with a mass flow rate of 6 kg/minat 1 MPa and 70 C and exits at 35 C. The cool-ing water enters at 300 kPa and 15 C andleaves at 25 C. Neglecting pressure drops,determine a) the required mass flow rate ofthe cooling water, and b) the heat transferrate from the refrigerant to the water.SolutionFirst consider the condenser as the control volume. The process is steady,adiabatic and no work is done. Thus over any time intervalΔt,ΔEΔt=0and thusXin˙E=Xout˙Ewhere˙E=˙m h+12V2+gz650:351 Thermodynamics·Prof. Doyle Knight37
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I only know this
sorry for errors
Answer:
to determine the rate of heat loss from that wall by convection = 12780 watts
Explanation:
