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:
V (initial vertical velocity) = 45.4 sin 31.2 = 23.52 m/s
1/2 m V^2 = m g h conservation of energy
h = V^2 / (2 g) = 23.52^2 / 19.6 = 28.2 m max height
Check:
t = 28.2 / 9.8 = 2.88 sec time to reach max height
h = 23.52 * 2.88 - 1/2 g 2.88^2 = 27.1 m
Zero.
Acceleration is defined as the change in velocity over time.
Since in your case there is no change, there is no acceleration, so it is zero:
Or in formula: <span>a=<span><span>Δv</span>t</span></span>
Where a=acceleration, <span>Δv</span>=change in velocity and t=time
The optimal angle of 45° for maximum horizontal range is only valid when initial height is the same as final height.
<span>In that particular situation, you can prove it like this: </span>
<span>initial velocity is Vo </span>
<span>launch angle is α </span>
<span>initial vertical velocity is </span>
<span>Vv = Vo×sin(α) </span>
<span>horizontal velocity is </span>
<span>Vh = Vo×cos(α) </span>
<span>total time in the air is the the time it needs to fall back to a height of 0 m, so </span>
<span>d = v×t + a×t²/2 </span>
<span>where </span>
<span>d = distance = 0 m </span>
<span>v = initial vertical velocity = Vv = Vo×sin(α) </span>
<span>t = time = ? </span>
<span>a = acceleration by gravity = g (= -9.8 m/s²) </span>
<span>so </span>
<span>0 = Vo×sin(α)×t + g×t²/2 </span>
<span>0 = (Vo×sin(α) + g×t/2)×t </span>
<span>t = 0 (obviously, the projectile is at height 0 m at time = 0s) </span>
<span>or </span>
<span>Vo×sin(α) + g×t/2 = 0 </span>
<span>t = -2×Vo×sin(α)/g </span>
<span>Now look at the horizontal range. </span>
<span>r = v × t </span>
<span>where </span>
<span>r = horizontal range = ? </span>
<span>v = horizontal velocity = Vh = Vo×cos(α) </span>
<span>t = time = -2×Vo×sin(α)/g </span>
<span>so </span>
<span>r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) </span>
<span>r = -(Vo)²×sin(2α)/g </span>
<span>To find the extreme values of r (minimum or maximum) with variable α, you must find the first derivative of r with respect to α, and set it equal to 0. </span>
<span>dr/dα = d[-(Vo)²×sin(2α)/g] / dα </span>
<span>dr/dα = -(Vo)²/g × d[sin(2α)] / dα </span>
<span>dr/dα = -(Vo)²/g × cos(2α) × d(2α) / dα </span>
<span>dr/dα = -2 × (Vo)² × cos(2α) / g </span>
<span>Vo and g are constants ≠ 0, so the only way for dr/dα to become 0 is when </span>
<span>cos(2α) = 0 </span>
<span>2α = 90° </span>
<span>α = 45° </span>
