The general accepted value of acceleration due to gravity, g, is 9.81 m/s^2.
That is an approximation because being the acceleration of gravity due to the attraction of the earth its magnitude will depend on the distance from the point to the center of the planet Earth.
The value of g is determined by using the Newton's Universal Law of gravity:
F = G * m of Earth * m of body / (distance^2)
Wehre {G* m of Earth / (distance^2) } = g
G is a universal constant = 6.67 * 10 ^ -11 N*m^2 / kg^2
m of Earth = 5.98 * 10 ^ 24 kg
distance = radius of Earth + height of the body
Given the the Earth is not a perfect sphere the radius varies. Also the height of the body varies.
If you take a mean radius of Earth of 6.37*10^6 m
you get
g = 6.67*10^-11 N*m^2/kg^2 * 5.98*10^24kg / (6.37*10^6 m)^2 = 9.83 m/s^2
Again, if you want a more precise value of g, you need to find the exact place where you are and then use the right r.
To solve the problem it is necessary to take into account the concepts related to frequency depending on the wavelength and the speed of light.
By definition we know that the frequency is equivalent to,
where,
c= Speed of light
While the wavelength is equal to,
Where,
L = Length
n = Number of antinodes/nodes
PART A) For the first part we have that our wavelength is 110MHz, therefore
Therefore the distance between the nodal planes is 1.36m
PART B) For this part we need to find the Length through the number of nodes (8) and the wavelength, that is,
Therefore the length of the cavity is 10.90m
D. is the correct answer.
Explanation:
initial height, yo = 2 m
initial velocity, u = 20 m/s
angle of projection,θ = 5 degree
distance of net = 7 m
height of net = 1 m
Let it covers a vertical distance y in time t .
Use Second equation of motion for vertical motion
As it hits the ground in time t, so put y = 0
Taking positive sign, t = 0.84 s
The ball travels a horizontal distance x in time t
X = 20 Cos5 x t
X = 16.76 m
As this distance is more than the distance of net, so it clears the net.
Let t' be the time taken to travel a horizontal distance equal to the distance of net
7 = 20 cos5 x t'
t' = 0.35 s
Let the vertical distance traveled by the ball in time t' is y'.
So,
y' = 2.008 m
So, it clears the net which is 1 m high.
It clears the net by a vertical distance of 2.008 - 1 = 1.008 m and horizontal distance 16.76 - 7 = 9.76 m
your welcome, and have a great day.
<span>This is the quantum mechanical model of the atom. In this model, electrons are represented by the probability that it lies somewhere in different energy level outside the nucleus. The foundation of quantum mechanics is the Schrodinger equation.</span>
