Answer:
They are used in imaging application gadgets such as video cameras,TV, surveillance cameras and document scanners
Explanation:
A charge couple device (CCDs) are highly capable in imagery detector.Its common application is in video and digital imaging.The quality of a charge couple device is determined by factors such as the dynamic range, dark charge level and the quantum efficiency.These devices serve the purpose of detecting optical images though some are installed with applications for data storage.
Answer:
Enthalpy of reaction (kJoules/mole)
Heat of formation of products (kJoules/mole)
Heat of reaction of reactants (kJoules/mole)
Explanation:
The general expression for calculating the overall enthalpy of reaction is given as following:
ΔH = ∑ΔH[producst] - ∑Δ[reactants]
Thus, the heat of reaction is given as the difference between the formation of the products and the formation of the reactants. The units are expressed as kJ/mol of reactants or products.
Thus, the three values are fundamental in the determination of the overall energy of the reaction from Hess' Law.
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
Answer:
radius = 0.045 m
Explanation:
Given data:
density of oil = 780 kg/m^3
velocity = 20 m/s
height = 25 m
Total energy is = 57.5 kW
we have now
E = kinetic energy+ potential energy + flow work
![E = \dot m ( \frac{v^2}{2] + zg + p\nu)](https://tex.z-dn.net/?f=E%20%3D%20%5Cdot%20m%20%28%20%5Cfrac%7Bv%5E2%7D%7B2%5D%20%2B%20%20zg%20%2B%20p%5Cnu%29)
![E = \dot m( \frac{v^2}{2] + zg + p_{atm} \frac{1}{\rho})](https://tex.z-dn.net/?f=E%20%3D%20%5Cdot%20m%28%20%5Cfrac%7Bv%5E2%7D%7B2%5D%20%2B%20%20zg%20%2B%20p_%7Batm%7D%20%5Cfrac%7B1%7D%7B%5Crho%7D%29)
solving for flow rate
![\dot m = 99.977we know that [tex]\dot m = \rho AV](https://tex.z-dn.net/?f=%5Cdot%20m%20%3D%2099.977%3C%2Fp%3E%3Cp%3Ewe%20know%20that%20%3C%2Fp%3E%3Cp%3E%5Btex%5D%5Cdot%20m%20%20%3D%20%5Crho%20AV)
solving for d
d = 0.090 m
so radius = 0.045 m
Answer:
The exit temperature is 293.74 K.
Explanation:
Given that
At inlet condition(1)
P =80 KPa
V=150 m/s
T=10 C
Exit area is 5 times the inlet area
Now
If consider that density of air is not changing from inlet to exit then by using continuity equation
So 
m/s
Now from first law for open system
Here Q=0 and w=0
When air is treating as ideal gas
Noe by putting the values
So the exit temperature is 293.74 K.
