Test questions measure recall; matching concepts with their definitions measures recognition.
<u>Explanation:
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According to Psychology our brain remembers everything what we learn but the understanding and remembering the right answer for the right question needs training and understanding ability. So in order to enhance the ability of recalling and recognizing among the students, the concept of test questions and matching with definitions are used in curricular activities.
As the students will be learning different terms, definitions, methods and different subjects, they should be able to distinguish among different definitions as well as they should recall the things they have learnt. So the answers for the test questions will help to recall the topics learnt by the students while the matching concept will help the students to recognize each definition with their terms.
Answer:
a. metallic bond
b. the valence electrons from the s and p orbitals of the interacting metal atoms delocalize. That is to say, instead of orbiting their respective metal atoms, they form a “cloud” of electrons that surrounds the positively charged atomic nuclei of the interacting metal ions.
c. due to the presence of free electrons in its outer energy levels
Answer:
Minimum thickness; t = 9.75 x 10^(-8) m
Explanation:
We are given;
Wavelength of light;λ = 585 nm = 585 x 10^(-9)m
Refractive index of benzene;n = 1.5
Now, let's calculate the wavelength of the film;
Wavelength of film;λ_film = Wavelength of light/Refractive index of benzene
Thus; λ_film = 585 x 10^(-9)/1.5
λ_film = 39 x 10^(-8) m
Now, to find the thickness, we'll use the formula;
2t = ½m(λ_film)
Where;
t is the thickness of the film
m is an integer which we will take as 1
Thus;
2t = ½ x 1 x 39 x 10^(-8)
2t = 19.5 x 10^(-8)
Divide both sides by 2 to give;
t = 9.75 x 10^(-8) m
Answer:
The solid material found in the centre of some planets at extremely high temperature and pressure, distinct from the liquid outer core.
about 1250 km
approximately 5700 K (5430 °C or 9806 °F)
Explanation:
Answer:
<em>The frequency of of the note = 131 Hz.</em>
Explanation:
<em>Frequency:</em><em> Frequency can be defined as the number of complete oscillation completed by a wave in one seconds. The S.I unit of frequency is Hertz ( Hz)</em>
v = λf ............................ Equation 1
Making f the subject of the equation,
f = v/λ .......................... Equation 2
Where v = Speed, λ = wavelength, f = frequency
<em>Given: v = 343 m/s, λ = 2.62 m.</em>
<em>Substituting these values into equation 2</em>
<em>f = 343/2.62</em>
<em>f = 131 Hz</em>
<em>Thus the frequency of of the note = 131 Hz.</em>
