Answer;
30.6 m
Explanation;
All objects accelerate at the constant rate in the Earth's gravitational field. The gravitational acceleration, g = 9.8 m/s².
Distance traveled by an object falling down under a constant acceleration will be given the formula;
s = ut² + 1/2(gt²); but u the initial velocity is o
thus;
S =1/2(gt²)
= 0.5 × 9.81 × 2.5 ²
= 30.65
≈ 30.6 m
Answer:
d= 0.242 mm
Explanation:
Slit width (d ) = ?
Screen distance ( D ) = 1.25 m
Wave length of light λ = 600 nm
Distance of n the dark fringe from centre
= n λ D / d
Here n = 2
so


d= 0.242 mm
Answer:
9.3 g/cm³
Explanation:
First, convert kg to g:
0.485 kg × (1000 g / kg) = 485 g
Density is mass divided by volume:
D = (485 g) / (52 cm³)
D = 9.33 g/cm³
Rounding to two significant figures, the density is 9.3 g/cm³.
Answer:

Explanation:
Given that,
Initially, the spaceship was at rest, u = 0
Final velocity of the spaceship, v = 11 m/s
Distance accelerated by the spaceship, d = 213 m
We need to find the acceleration experienced by the occupants of the spaceship during the launch. It is a concept based on the equation of kinematics. Using the third equation of motion to find acceleration.

So, the acceleration experienced by the occupants of the spaceship is
.
8 wave units I guess I tried it should be the answer though