it is true. the trajectory reaches the value of zero at the top
Linear expansivity, area expansivity and volume or cubic expansivity are
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Answer:
Maximum altitude to see(L) = 1.47 × 10⁶ m (Approx)
Explanation:
Given:
wavelength (λ) = 0.12 nm = 0.12 × 10⁻⁹ m
Pupil Diameter (d) = 4.1 mm = 4 × 10⁻³ m
Separation distance (D) = 5.4 cm = 0.054 m
Find:
Maximum altitude to see(L)
Computation:
Resolving power = 1.22(λ / d)
D / L = 1.22(λ / d)
0.054 / L = 1.22 [(0.12 × 10⁻⁹) / (4 × 10⁻³ m)]
0.054 / L = 1.22 [0.03 × 10⁻⁶]
L = 0.054 / 1.22 [0.03 × 10⁻⁶]
L = 0.054 / [0.0366 × 10⁻⁶]
L = 1.47 × 10⁶
Maximum altitude to see(L) = 1.47 × 10⁶ m (Approx)
<span>Mass of the block m = 3.3kg
Angle of the slide = 30 degrees
Distance the block slides s = 2.10 m
Time taken to slide t = 1.6 s
Initially in rest condition so initial velocity u = 0.
We have an equation for distance s = (u x t) + (1/2) x (a t^2)
s = (0 x t) + (1/2) x (a x (1.6) ^2) => 2.10 = (1/2) x (a x2.56)
2.56a = 4.20 => a = 1.64
So the magnitude of the Acceleration a = 1.64 m/s^2</span>
