Answer:
θ = 22.2
Explanation:
This is a diffraction exercise
a sin θ = m λ
The extension of the third zero is requested (m = 3)
They indicate the wavelength λ = 630 nm = 630 10⁻⁹ m and the width of the slit a = 5 10⁻⁶ m
sin θ = m λ / a
sin θ = 3 630 10⁻⁹ / 5 10⁻⁶
sin θ = 3.78 10⁻¹ = 0.378
θ = sin⁻¹ 0.378
to better see the result let's find the angle in radians
θ = 0.3876 rad
let's reduce to degrees
θ = 0.3876 rad (180º /π rad)
θ = 22.2º
Answer:
x ’= 368.61 m, y ’= 258.11 m
Explanation:
To solve this problem we must find the projections of the point on the new vectors of the rotated system θ = 35º
x’= R cos 35
y’= R sin 35
The modulus vector can be found using the Pythagorean theorem
R² = x² + y²
R = 450 m
we calculate
x ’= 450 cos 35
x ’= 368.61 m
y ’= 450 sin 35
y ’= 258.11 m
PE = mass * height * 9.81
PE = 142 * 25 * 9.81
PE = 34825.5 J
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