Answer:
Coiled tubing is often used to carry out operations similar to wire lining.
Answer:
450,000m = 450km = 4.5E5
32,600,000W = 32.6MW = 3.26E7
59,700,000,000cal = 59.7Gcal = 5.97E10
0.000000083s = 83ns = 8.3E-8
35,000Ω = 35kΩ = 3.5E4
Explanation:
Giga = 1,000,000,000
Mega = 1,000,000
kilo = 1,000
unit = 1
deci = .1
centi = .01
milli = .001
micro = .000001
nano = .0000000001
pico = .000000000001
You should be able to look at these and convert between them in seconds if you want to pursue anything in engineering.
Answer:
If Reynolds number increases the extent of the region around the object that is affected by viscosity decreases.
Explanation:
Reynolds number is an important dimensionless parameter in fluid mechanics.
It is calculated as;
where;
ρ is density
v is velocity
d is diameter
μ is viscosity
All these parameters are important in calculating Reynolds number and understanding of fluid flow over an object.
In aerodynamics, the higher the Reynolds number, the lesser the viscosity plays a role in the flow around the airfoil. As Reynolds number increases, the boundary layer gets thinner, which results in a lower drag. Or simply put, if Reynolds number increases the extent of the region around the object that is affected by viscosity decreases.
Senors are a type of device that produce a amount of change to the output to a known input stimulus.
Input signals are signals that receive data by the system and outputs the ones who are sent from it. Hope this helps ;)
Answer:
work is 50 kj
Explanation:
Given data
heat (Q) = 50 kj
To find out
work input for the compression stroke per kilogram of air
Solution
we will apply here "first law of thermodynamics" i.e.
The First Law of Thermodynamics states that heat is a form of energy, subject to the principle of conservation of energy, that heat energy cannot be created or destroyed. It can be transferred from one location to another location. i.e.
ΔU = Q – W ................1
here ΔU is change in internal energy, Q is heat and W is work done
here U = 0 because air compressor the compression takes place at a constant internal energy in question
so that by equation 1
Q = W
and Q = 50
so work will be 50 kj
