Answer:
Wavelength, frequency and the photon energy changes as the one goes across the ranges of the electro-magnetic radiations.
Explanation:
Electro-magnetic radiations may be defined as the form of energy that is radiated or given by the electro-magnetic radiations. The visible light that we can see is the one of the electro-magnetic radiations. Other forms are the radio waves, gamma waves, UV rays, infrared radiations, etc.
The wavelength of the radiations decreases as we go from a. radio waves -- b. infrared radiation -- c. visible light -- d. ultraviolet radiation -- e. gamma radiation.
The frequency of the radiations increases when we move from a. radio waves -- b. infrared radiation -- c. visible light -- d. ultraviolet radiation -- e. gamma radiation.
The photon energy of the radiations increases when we move from a. radio waves -- b. infrared radiation -- c. visible light -- d. ultraviolet radiation -- e. gamma radiation.
Answer:
1.52 seconds
Explanation:
Step 1: identity the given parameters
Initial velocity (u) = 12m/s
Height above ground (h1) = 4m
Final velocity (V) = 0
Step 2: calculate the height travelled by the object from 4m height (h2).
V^2 = U^2 -2gh
0= 12^2-2(9.8*h)
2(9.8*h) = 12^2
19.6*h = 144
h = 144/19.6
h = 7.347 m
Total height above ground (ht) = 4m +7.347m = 11.347m
Step 3: calculate the time reach ground
T = √(2h/g)
T = √(2*11.347/9.8)
T= √(22.694/9.8)
T= √2.316
T= 1.52 seconds
Answer:
a) , b) , c)
Explanation:
a) The turbine is modelled by means of the First Principle of Thermodynamics. Changes in kinetic and potential energy are negligible.
The mass flow rate is:
According to property water tables, specific enthalpies and entropies are:
State 1 - Superheated steam
State 2s - Liquid-Vapor Mixture
The isentropic efficiency is given by the following expression:
The real specific enthalpy at outlet is:
State 2 - Superheated Vapor
The mass flow rate is:
b) The temperature at the turbine exit is:
c) The rate of entropy generation is determined by means of the Second Law of Thermodynamics:
Answer:
(a) 13.6 eV
(b) 10.2 V
Explanation:
a) Ionization potential energy is defined as the minimum energy required to excite a neutral atom to its ionized state i.e basically the minimum energy required to excite an electron from n=1 to infinity.
Energy of a level, n, in Hydrogen atom is,
Now ionization potential can be calculated as
Substitute all the value of energy and n in above equation.
Therefore, the ionization potential is 13.6 eV.
b) This is the energy required to excite a atom from ground state to its excited state. When electrons jumps from ground state level(n=1) to 1st excited state(n=2) the corresponding energy is called 1st excitation potential energy and corresponding potential is called 1st excitation potential.
So, 1st excitation energy = E(n 2)- E(n = 1)
Now we can find that 1st excitation energy is 10.2 eV which gives,
Therefore, the 1st excitation potential is 10.2V.
Answer:
Application engineering involves the production or creation of the devices or products by the implementation of technology. Application engineers usually design, create ,test and modify their products etc. So using glue to replace metal fasteners in electronics and automotive products best exemplifies <u>Application Engineering</u><u>.</u>