Answer:
D. F = 24000[N]
Explanation:
To be able to calculate the force we must first find the pressure at the bottom of the tank. By means of the following equation:
where:
Ro = density of the liquid = 800 [kg/m³]
g = gravity acceleration = 9.81 [m/s²]
h = elevation = 1.2 [m]
![P=800*9.81*1.2\\P=9417.6[Pa]](https://tex.z-dn.net/?f=P%3D800%2A9.81%2A1.2%5C%5CP%3D9417.6%5BPa%5D)
Pressure is defined as the relationship between Force on the area.
![P=F/A\\F=9417.6*2.5\\F = 23544 [N] = 24000[N]](https://tex.z-dn.net/?f=P%3DF%2FA%5C%5CF%3D9417.6%2A2.5%5C%5CF%20%3D%2023544%20%5BN%5D%20%3D%2024000%5BN%5D)
We need it to calculate the how big it is, it’s mass, or even volume to get the measurements of lots of objects
Answer:
The thermal conductivity of the wall = 40W/m.C
h = 10 W/m^2.C
Explanation:
The heat conduction equation is given by:
d^2T/ dx^2 + egen/ K = 0
The thermal conductivity of the wall can be calculated using:
K = egen/ 2a = 800/2×10
K = 800/20 = 40W/m.C
Applying energy balance at the wall surface
"qL = "qconv
-K = (dT/dx)L = h (TL - Tinfinity)
The convention heat transfer coefficient will be:
h = -k × (-2aL)/ (TL - Tinfinty)
h = ( 2× 40 × 10 × 0.05) / (30-26)
h = 40/4 = 10W/m^2.C
From the given temperature distribution
t(x) = 10 (L^2-X^2) + 30 = 30°
T(L) = ( L^2- L^2) + 30 = 30°
dT/ dx = -2aL
d^2T/ dx^2 = - 2a
Answer:
A) When the angle between the Force (F) and Displacement (x) is 0°, because, Work done (W) is directly proportional to the Cosine of the Angle between the Force applied and the resultant displacement of the subject.
W = F•x cos ∅
If ∅ = 0°,
W = F•x ===> Maximum Work Done.
If ∅ = 45°,
W = F•x/√2
If ∅ = 90°,
W = 0
If ∅ = 180°,
W = –F•x ===> Minimum Work Done.
