Answer:
Approaches mathematical learning through inquiry
-Explore real contexts, problems, situations, and models
-Learning through doing shifts the focus on the students
-Problems have multiple entry and exit points
-Links to other disciplines
Explanation:
quizlet
<span>Answer a. is correct. In this case the dog treats were positive reinforcers for staying close (the desirable behaviour). Repeated rewards for a certain behaviour work as positive reinforcers.</span>
Answer:
1/4 λ film
Explanation:
Let L = total path length then L = 2 t where t is film thickness
There will be a 180 deg phase change at the air-film interface but no
phase change at the film-air interface
L = n * wavelength / 2 where n = 1, 3, 5, 7 etc
(the total path L must be an odd number of 1/2 wavelengths)
Or t = n * wavelength / 4 (the film must be an odd number
of 1/4 wavelengths thick)
1/4 λ film satisfies this condition
Note: Find the missing diagram in the attachment section.
Answer:
The value of the angle is ![\bf{ \sin^{-1}[h/am_{e}v]}](https://tex.z-dn.net/?f=%5Cbf%7B%20%5Csin%5E%7B-1%7D%5Bh%2Fam_%7Be%7Dv%5D%7D)
.
Explanation:
Given:
The condition for diffraction minima is
where, 
is the slit-width, 
is the angle of incidence, 
is the order number and 
is the wavelength of the light.
The wavelength of an electron traveling through a medium is governed by de Broglie's hypothesis.
According to de Broglie's hypothesis
Here, 
is Planck's constant, 
is the mass of the electron and 
is the velocity of the electron.
For first minimum 
.
From equation (1), we have
![&& a \sin \theta = \dfrac{h}{m_{e}v}\\&or,& \theta = \sin^{-1}[\dfrac{h}{am_{e}v}]](https://tex.z-dn.net/?f=%26%26%20a%20%5Csin%20%5Ctheta%20%3D%20%5Cdfrac%7Bh%7D%7Bm_%7Be%7Dv%7D%5C%5C%26or%2C%26%20%5Ctheta%20%3D%20%5Csin%5E%7B-1%7D%5B%5Cdfrac%7Bh%7D%7Bam_%7Be%7Dv%7D%5D)
Answer:
Explanation:
Given that,
Power of electromagnetic energy, P = 100 W
We need to find the intensity at a distance of 10 m from the source. Intensity is equal to the power per unit area. So,
So, the intensity at a distance of 10 m is 
