Answer:
r = 2.031 x 10⁶ m = 2031 km
Explanation:
In order for the asteroid to orbit the planet, the centripetal force must be equal to the gravitational force between asteroid and planet:
Centripetal Force = Gravitational Force
mv²/r = GmM/r²
v² = GM/r
r = GM/v²
where,
r = radial distance = ?
G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
M = Mass of Planet = 3.52 x 10¹³ kg
v = tangential speed = 0.034 m/s
Therefore,
r = (6.67 x 10⁻¹¹ N.m²/kg²)(3.52 x 10¹³ kg)/(0.034 m/s)²
<u>r = 2.031 x 10⁶ m = 2031 km</u>
Answer:
v = 8.65 m/s
Explanation:
Given that,
Distance covered by the doge, d = 45 m
Time taken, t = 5.2 s
We need to find its average speed. The total distance covered divided by the total time taken is called the average speed of an object. So,
So, the average speed is 8.65 m/s.
Answer:
Explanation:
The question relates to time of flight of a projectile .
Time of flight = 2 u sinθ / g
u is speed of projectile , θ is angle of projectile
= 2 x 48.5 sin42 / 9.8
= 6.6 seconds .
Maximum height attained
= u² sin²θ / g
= 48.5² sin²42 / 9.8
= 107.47 m .
Answer:
1.696 nm
Explanation:
For a diffraction grating, dsinθ = mλ where d = number of lines per metre of grating = 5510 lines per cm = 551000 lines per metre and λ = wavelength of light = 467 nm = 467 × 10⁻⁹ m. For a principal maximum, m = 1. So,
dsinθ = mλ = (1)λ = λ
dsinθ = λ
sinθ = λ/d.
Also tanθ = w/D where w = distance of center of screen to principal maximum and D = distance of grating to screen = 1.03 m
From trig ratios 1 + cot²θ = cosec²θ
1 + (1/tan²θ) = 1/(sin²θ)
substituting the values of sinθ and tanθ we have
1 + (D/w)² = (d/λ)²
(D/w)² = (d/λ)² - 1
(w/D)² = 1/[(d/λ)² - 1]
(w/D) = 1/√[(d/λ)² - 1]
w = D/√[(d/λ)² - 1] = 1.03 m/√[(551000/467 × 10⁻⁹ )² - 1] = 1.03 m/√[(1179.87 × 10⁹ )² - 1] = 1.03 m/1179.87 × 10⁹ = 0.000848 × 10⁻⁹ = 0.848 × 10⁻¹² m = 0.848 nm.
w is also the distance from the center to the other principal maximum on the other side.
So for both principal maxima to be on the screen, its minimum width must be 2w = 2 × 0.848 nm = 1.696 nm
So, the minimum width of the screen must be 1.696 nm
