Answer:
The expression is shown in the explanation below:
Explanation:
Thinking process:
Let the time period of a simple pendulum be given by the expression:

Let the fundamental units be mass= M, time = t, length = L
Then the equation will be in the form


where k is the constant of proportionality.
Now putting the dimensional formula:
![T = KM^{a}L^{b} [LT^{-} ^{2}]^{c}](https://tex.z-dn.net/?f=T%20%3D%20KM%5E%7Ba%7DL%5E%7Bb%7D%20%20%5BLT%5E%7B-%7D%20%5E%7B2%7D%5D%5E%7Bc%7D)

Equating the powers gives:
a = 0
b + c = 0
2c = 1, c = -1/2
b = 1/2
so;
a = 0 , b = 1/2 , c = -1/2
Therefore:

T = 
where k = 
Answer:
The differential equation and the boundary conditions are;
A) -kdT(r1)/dr = h[T∞ - T(r1)]
B) -kdT(r2)/dr = q'_s = 734.56 W/m²
Explanation:
We are given;
T∞ = 70°C.
Inner radii pipe; r1 = 6cm = 0.06 m
Outer radii of pipe;r2 = 6.5cm=0.065 m
Electrical heat power; Q'_s = 300 W
Since power is 300 W per metre length, then; L = 1 m
Now, to the heat flux at the surface of the wire is given by the formula;
q'_s = Q'_s/A
Where A is area = 2πrL
We'll use r2 = 0.065 m
A = 2π(0.065) × 1 = 0.13π
Thus;
q'_s = 300/0.13π
q'_s = 734.56 W/m²
The differential equation and the boundary conditions are;
A) -kdT(r1)/dr = h[T∞ - T(r1)]
B) -kdT(r2)/dr = q'_s = 734.56 W/m²
Answer:

Explanation:
The power needed to make the escalator working is obtained by means of the Work-Energy Theorem:




The mechanical efficiency of the escalator is:

