Answer:
The electric ceiling fan was invented in 1882 by engineer and inventor, Philip Diehl. He had earlier invented an electric sewing machine and adapted the motor from this invention to create the ceiling fan. He called his invention the “Diehl Electric Fan” and it was such a success that he soon had many other people competing with him.
Explanation:
Answer:
Only Technician B is right.
Explanation:
The cylindrical braking system for a car works through the mode of pressure transmission, that is, the pressure applied to the brake pedals, is transmitted to the brake pad through the cylindrical piston.
Pressure applied on the pedal, P(pedal) = P(pad)
And the Pressure is the applied force/area for either pad or pedal. That is, P(pad) = Force(pad)/A(pad) & P(pedal) = F(pedal)/A(pedal)
If the area of piston increases, A(pad) increases and the P(pad) drops, Meaning, the pressure transmitted to the pad reduces. And for most cars, there's a pressure limit for the braking system to work.
If the A(pad) increases, P(pad) decreases and the braking force applied has to increase, to counter balance the dropping pressure and raise it.
This whole setup does not depend on the length of the braking lines; it only depends on the applied force and cross sectional Area (size) of the piston.
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:
1700kJ/h.K
944.4kJ/h.R
944.4kJ/h.°F
Explanation:
Conversions for different temperature units are below:
1K = 1°C + 273K
1R = T(K) * 1.8
= (1°C + 273) * 1.8
1°F = (1°C * 1.8) + 32
Q/delta T = 1700kJ/h.°C
T (K) = 1700kJ/h.°C
= 1700kJ/K
T (R) = 1700kJ/h.°C
= 1700kJ/h.°C * 1°C/1.8R
= 944.4kJ/h.R
T (°F) = 1700kJ/h.°C
= 1700kJ/h.°C * 1°C/1.8°F
= 944.4kJ/h.°F
Note that arithmetic operations like subtraction and addition of values do not change or affect the value of a change in temperature (delta T) hence, the arithmetic operations are not reflected in the conversion. Illustration: 5°C - 3°C
= 2°C
(273+5) - (273+3)
= 2 K
That would be B, I hope this helps!
