The solution would be like this for this specific problem:
Given:
diffraction grating
slits = 900 slits per centimeter
interference pattern that
is observed on a screen from the grating = 2.38m
maxima for two different
wavelengths = 3.40mm
slit separation .. d =
1/900cm = 1.11^-3cm = 1.111^-5 m <span>
Whenas n = 1, maxima (grating equation) sinθ = λ/d
Grant distance of each maxima from centre = y ..
<span>As sinθ ≈ y/D y/D =
λ/d λ = yd / D </span>
∆λ = (λ2 - λ1) = y2.d/D - y1.d/D
∆λ = (d/D) [y2 -y1]
<span>∆λ = 1.111^-5m x [3.40^-3m] / 2.38m .. .. ►∆λ = 1.587^-8 m</span></span>
Answer:
a. A = 0.1656 m
b. % E = 1.219
Explanation:
Given
mB = 4.0 kg , mb = 50.0 g = 0.05 kg , u₁ = 150 m/s , k = 500 N / m
a.
To find the amplitude of the resulting SHM using conserver energy
ΔKe + ΔUg + ΔUs = 0
¹/₂ * m * v² - ¹/₂ * k * A² = 0
A = √ mB * vₓ² / k
vₓ = mb * u₁ / mb + mB
vₓ = 0.05 kg * 150 m / s / [0.050 + 4.0 ] kg = 1.8518
A = √ 4.0 kg * (1.852 m/s)² / (500 N / m)
A = 0.1656 m
b.
The percentage of kinetic energy
%E = Es / Ek
Es = ¹/₂ * k * A² = 500 N / m * 0.1656²m = 13.72 N*0.5
Ek = ¹/₂ * mb * v² = 0.05 kg * 150² m/s = 1125 N
% E = 13.72 / 1125 = 0.01219 *100
% E = 1.219
Answer:
The fence is 5feet less.
Explanation:
We need to determine
The less amount of fence required, if the enclosure has full width and reduced length, compared to full length and reduced width.
Approach & WorkingArea of lawn = 30 × 403/4th of the area of lawn = ¾(30 × 40) = 30 * 30
When full width will be fenced, and reduced length will be fenced.
Width = 30 feet30 * L = 30 * 30Hence, length = 30 feetLength of fence needed = 2(30 + 30) = 120 feet
When full length will be fenced, and reduced width will be fenced
Length = 40 feet40 * W = 30 * 30W = 22.5 feetLength of fence needed = 2(40 + 22.5) = 125 feet
Difference in length of fence needed = 125 – 120 = 5 feet.
Transform boundary
is the answer from stemscopes
