Answer:
6.18 um
Explanation:
The plumb line will be pulled down by a combination of the gravitationall pull of Earth and of the mountain. The Earth pulls down and the mountain to the side. Because of this it will fall not in a straight line down, but slightly to the side. Since the plumb line will follow the compound gravity we can imagine a rectangle triangle formed by the plumb line, a vertical line that ends at the same height as the plumb line, and the sideways displacement.
The total gravity will be proportional to the plumb line lenght, the vertical line will be proportional to Earth's gravity and the sideways displacement to the mountain pull.
The gravity of Earth is 9.81 m/s^2
The pull of the mountain will be defined by Newton's law of universal gravitation:
Where
F: pull force
G: universal gravitational constant (6.67e-11 m^3/(kg * s)
m1: mass of the mountain
m2: mass of the plumb
r: distance between mountain and plumb (3 km in this case)
If we divide both sides by m2 we obtain the acceleration towards the mountain of the plumb
Now we need the mass of the mountain. This will be its volume times it's density. The volume depends on the radius (since we consider it as a sphere)
So, the acceleration on the plumb will be
This is very small compared to the pull of Earth, so we can make an approximation that the length of the plumb line is equal to vertical line.
We can use the principle of similar triangles to say that:
So:
V^2-u^2=2as
v=final velocity=unkown
u=initial velocity=0 m/s, because freely falling
a=acceleration due to gravity=9.8 m/s^2
s=distance (here height) traveled=4.5m
therefore the final velocity,
v^2=2*9.8*4.5
v=<span>9.39m/s</span>
Answer:
mass of the planet and its distance from other objects
Explanation:
mass of the planet and its distance from other objects determine the strength of its gravitational force on other objects
The wavelength of the light that are not reflected in the air is,
λ5= 600 nm, λ6= 500 nm, λ7=428 nm.
<u>Explanation:</u>
- For the above problem, we have the case as no<n1 and n2 < n1 .
- so we have(1<1.5;1.4<1.5)To find the wavelengths of the light that is not reflected in the air.
- Since the phase changes are not common on both the surfaces/
- we have the formula, λ=2d n²/m ( where m=1,2,3,4,5...)
- solving the problem we get,
- λ5= 600 nm, λ6= 500 nm, λ7=428 nm
- The wavelength of the light that are not reflected in the air is,
λ5= 600 nm, λ6= 500 nm, λ7=428 nm.