Answer:
<em>The 6000 lines per cm grating, will produces the greater dispersion .</em>
Explanation:
A diffraction grating is an optical component with a periodic (usually one that has ridges or rulings on their surface rather than dark lines) structure that splits and diffracts light into several beams travelling in different directions.
The directions of the light beam produced from a diffraction grating depend on the spacing of the grating, and also on the wavelength of the light.
For a plane diffraction grating, the angular positions of principle maxima is given by
(a + b) sin ∅n = nλ
where
a+b is the distance between two consecutive slits
n is the order of principal maxima
λ is the wavelength of the light
From the equation, we can see that without sin ∅ exceeding 1, increasing the number of lines per cm will lead to a decrease between the spacing between consecutive slits.
In this case, light of the same wavelength is used. If λ and n is held constant, then we'll see that reducing the distance between two consecutive slits (a + b) will lead to an increase in the angle of dispersion sin ∅. So long as the limit of sin ∅ not greater that one is maintained.
Answer:
Approximately 
(given that the magnitude of this charge is 
.)
Explanation:
If a charge of magnitude 
is placed in an electric field of magnitude 
, the magnitude of the electrostatic force on that charge would be 
.
The magnitude of this charge is 
. Apply the unit conversion 
:
.
An electric field of magnitude 
would exert on this charge a force with a magnitude of:
.
Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.
Imagine a car, and imagine you see little arrows below and beneath it, just around it. <span>The path of a particle that is flowing steadily and without turbulence in a fluid past an object.</span>
Answer:
The peak-to-peak ripple voltage = 2V
Explanation:
120V and 60 Hz is the input of an unfiltered full-wave rectifier
Peak value of output voltage = 15V
load connected = 1.0kV
dc output voltage = 14V
dc value of the output voltage of capacitor-input filter
where
V(dc value of output voltage) represent V₀
V(peak value of output voltage) represent V₁
V₀ = 1 - ( 
)V₁
make C the subject of formula
V₀/V₁ = 1 - (1 / 2fRC)
1 / 2fRC = 1 - (v₀/V₁)
C = 2fR ((1 - (v₀/V₁))⁻¹
Substitute for,
f = 240Hz , R = 1.0Ω, V₀ = 14V , V₁ = 15V
C = 2 * 240 * 1 (( 1 - (14/15))⁻¹
C = 62.2μf
The peak-to-peak ripple voltage
= (1 / fRC)V₁
= 1 / ( (120 * 1 * 62.2) )15V
= 2V
The peak-to-peak ripple voltage = 2V
