After rolling off the edge of the cliff and falling ' M ' meters down,
the speed of the boulder is
Square root of ( 19.6 M ) .
If M=111 meters, then the speed is <em>46.64 meters per second</em>.
We have known for roughly 500 years that if there's no air resistance,
the mass of the falling object makes no difference, and all objects fall
with the same acceleration, speed, time to splat, etc.
Answer : The energy of one photon of hydrogen atom is, 
Explanation :
First we have to calculate the wavelength of hydrogen atom.
Using Rydberg's Equation:

Where,
= Wavelength of radiation
= Rydberg's Constant = 10973731.6 m⁻¹
= Higher energy level = 3
= Lower energy level = 2
Putting the values, in above equation, we get:


Now we have to calculate the energy.

where,
h = Planck's constant = 
c = speed of light = 
= wavelength = 
Putting the values, in this formula, we get:


Therefore, the energy of one photon of hydrogen atom is, 
which of the following is not a barrier to physical activity it is fear of injury I think.
They are falling under the sole influence of gravity all objects<span> will </span>fall<span> with the </span>same<span> rate of </span><span>acceleration needless of there size</span>
Answer:

Explanation:
The force on the point charge q exerted by the rod can be found by Coulomb's Law.

Unfortunately, Coulomb's Law is valid for points charges only, and the rod is not a point charge.
In this case, we have to choose an infinitesimal portion on the rod, which is basically a point, and calculate the force exerted by this point, then integrate this small force (dF) over the entire rod.
We will choose an infinitesimal portion from a distance 'x' from the origin, and the length of this portion will be denoted as 'dx'. The charge of this small portion will be 'dq'.
Applying Coulomb's Law:

The direction of the force on 'q' is to the right, since both charges are positive, and they repel each other.
Now, we have to write 'dq' in term of the known quantities.

Now, substitute this into 'dF':

Now we can integrate dF over the rod.
