Nothing flammable of explosive type of material is around
Answer:
R= 1.25
Explanation:
As given the local heat transfer,

But we know as well that,

Replacing the values

Reynolds number is define as,

Where V is the velocity of the fluid and \upsilon is the Kinematic viscosity
Then replacing we have



<em>*Note that A is just a 'summary' of all of that constat there.</em>
<em>That is
</em>
Therefore at x=L the local convection heat transfer coefficient is

Definen that we need to find the average convection heat transfer coefficient in the entire plate lenght, so

The ratio of the average heat transfer coefficient over the entire plate to the local convection heat transfer coefficient is

Answer:
Not possible.
Explanation:
According to second law of thermodynamics, the maximum efficiency any heat engine could achieve is Carnot Efficiency η defined by:

Where
and
are temperature (in Kelvin) of heat source and heatsink respectively
In our case (I will be using K = 273+°C) :

In percentage, this is 14.28% efficiency, which is the <em>maximum</em> theoretical efficiency <em>any</em> heat engine could have while working between -27 and 14 °C temperature. Any claim of more efficient heat engine between these 2 temperature are violates the second law of thermodynamics. Therefore, the claim must be false.
Answer:
The resultant moment is 477.84 N·m
Explanation:
We note that the resultant moment is given by the moment about a given point
The length of the sides of the formed triangles are;
l = sin(40°) × 4/sin(110°) ≈ 2.736
Taking the moment about the lower left hand corner of the figure, with the convention that clockwise moments are positive, we have;
The resultant moment, ∑m, is given as follow;
∑M = 250 N × 4 m + 400 N × cos(40°) × 4 m - 400 N × cos(40°) × 2 m + 400 N × sin(40°) × 2 m × tan(40°) - 600 N × cos(40°) × 2 m - 600 N× sin(40°) × 2 m × tan(40°) = 477.837084 N·m
Therefore, the resultant moment, ∑m ≈ 477.84 N·m clockwise.