Answer:
True The grid with more slits gives more angle separation increases
True. The grating with 10 slits produces better-defined (narrower) peaks
Explanation:
Such a system can be seen as a diffraction network in this case with different number of lines per unit length, the expression for the constructive interference of a diffraction network is
d sin θ = m λ
where d is the distance between slits or lines, m the order of diffraction and λ the wavelength.
For network with 5 slits
d = 1/5 = 0.2
For the network with 10 slits
d = 1/10 = 0.1
let's calculate the separation (teat) for each one
θ = sin⁻¹ (m λ / d)
for 5 slits
θ₅ = sin⁻¹ (m λ 5)
for 10 slits
θ₁₀ = sin⁻¹ (m λ 10)
we can appreciate that for more slits the angle increases
the intensity of a series of slits is
I = I₀ sin²2 (N d/2) / sin² d/2)
when there are more slits (N) the peaks have greater intensity and are more acute (half width decreases)
let's analyze the claims
False
True The grid with more slits gives more angle separation increases
False
True The expression for the intensity of the diffraction peaks the intensity of the peaks increases with the number of slits as well as their spectral width decreases
False
Answer:
A) μ = A.m²
B) z = 0.46m
Explanation:
A) Magnetic dipole moment of a coil is given by; μ = NIA
Where;
N is number of turns of coil
I is current in wire
A is area
We are given
N = 300 turns; I = 4A ; d =5cm = 0.05m
Area = πd²/4 = π(0.05)²/4 = 0.001963
So,
μ = 300 x 4 x 0.001963 = 2.36 A.m².
B) The magnetic field at a distance z along the coils perpendicular central axis is parallel to the axis and is given by;
B = (μ_o•μ)/(2π•z³)
Let's make z the subject ;
z = [(μ_o•μ)/(2π•B)] ^(⅓)
Where u_o is vacuum permiability with a value of 4π x 10^(-7) H
Also, B = 5 mT = 5 x 10^(-6) T
Thus,
z = [ (4π x 10^(-7)•2.36)/(2π•5 x 10^(-6))]^(⅓)
Solving this gives; z = 0.46m =
Answer: set up proportions
Explanation:
Answer:
the image is behind the mirror
virtual
erect(not inverted)
larger than the object
Answer:
Correct answer: E total = 2,800 J
Explanation:
Given:
m = 4 kg the mass of the object
V = 20 m/s the speed (velocity) of the object
H = 50 m the height of the object above the surface
E total = ? J
The total energy of an object is equal to the sum of potential and kinetic energy
E total = Ep + Ek
Ep = m g H we take g = 10 m/s²
Ep = 4 · 10 · 50 = 2,000 J
Ek = m V² / 2
Ek = 4 · 20² / 2 = 2 · 400 = 800 J
E total = 2,000 + 800 = 2,800 J
E total = 2,800 J
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