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Answer:
Explanation:
given
T = 3months = 7.9 × 10⁶s
orbital speed = 88 × 10³m/s
V= 2πr÷T
∴ r = (V×T) ÷ 2π
r = (88km × 7.9 × 10⁶s) ÷ 2π
r = 1.10 × 10⁸km
using kepler's 3rd law
mass of both stars = (seperation diatance)³/(orbital speed)²
M₁ + M₂ = (2r)³/(year)²
= (1.06 × 10²⁵)/(6.2×10¹³)
1.71×10¹²kg
since M₁ = M₂ =1.71×10¹²kg ÷ 2
M₁ = M₂ = 8.55×10¹¹kg
Answer:
<h2> 1.643*10⁻⁴cm</h2>Explanation:
In a single slit experiment, the distance on a screen from the centre point is expressed as y = where;
is the first two diffraction minima = 1
is light wavelength
d is the distance of diffraction pattern from the screen
a is the width of the slit
Given = 460-nm = 460*10⁻⁹m
d = 5.0mm = 5*10⁻³m
a = 1.4mm = 1.4*10⁻³m
Substituting this values into the formula above to get width of the central maximum y;
y = 1*460*10⁻⁹ * 5*10⁻³/1.4*10⁻³
y = 2300*10⁻¹²/1.4*10⁻³
y = 1642.86*10⁻⁹
y = 1.643*10⁻⁶m
Converting the final value to cm,
since 100cm = 1m
x = 1.643*10⁻⁶m
x = 1.643*10⁻⁶ * 100
x = 1.643*10⁻⁴cm
Hence, the width of the central maximum in the diffraction pattern on a screen 5.0 mm away is 1.643*10⁻⁴cm