For a neutral atom number of electrons equals number of protons, in other for the net charge of the atom to be zero...
no. of electrons = 12
Well, I guess you can come close, but you can't tell exactly.
It must be presumed that the seagull was flying through the air
when it "let fly" so to speak, so the jettisoned load of ballast
of which the bird unburdened itself had some initial horizontal
velocity.
That impact velocity of 98.5 m/s is actually the resultant of
the horizontal component ... unchanged since the package
was dispatched ... and the vertical component, which grew
all the way down in accordance with the behavior of gravity.
98.5 m/s = √ [ (horizontal component)² + (vertical component)² ].
The vertical component is easy; that's (9.8 m/s²) x (drop time).
Since we're looking for the altitude of launch, we can use the
formula for 'free-fall distance' as a function of acceleration and
time:
Height = (1/2) (acceleration) (time²) .
If the impact velocity were comprised solely of its vertical
component, then the solution to the problem would be a
piece-o-cake.
Time = (98.5 m/s) / (9.81 m/s²) = 10.04 seconds
whence
Height = (1/2) (9.81) (10.04)²
= (4.905 m/s²) x (100.8 sec²) = 494.43 meters.
As noted, this solution applies only if the gull were hovering with
no horizontal velocity, taking careful aim, and with malice in its
primitive brain, launching a remote attack on the rich American.
If the gull was flying at the time ... a reasonable assumption ... then
some part of the impact velocity was a horizontal component. That
implies that the vertical component is something less than 98.5 m/s,
and that the attack was launched from an altitude less than 494 m.
Answer:
a
The mass is 
b
The velocity is 
Explanation:
From the question we are told that
The speed of the protons is 
The mass of the protons is 
The speed of the rebounding protons are 
The negative sign shows that it is moving in the opposite direction
Now according to the law of energy conservation mass of one nucleus of the unknown element. is mathematically represented as
![m_2 = [\frac{u_1 -v_1}{u_1 + v_1} ] m_1](https://tex.z-dn.net/?f=m_2%20%3D%20%5B%5Cfrac%7Bu_1%20-v_1%7D%7Bu_1%20%2B%20v_1%7D%20%5D%20m_1)
Where 
is the mass of a single proton
So substituting values
The mass of on proton is 
So 
Now according to the law of linear momentum conservation the speed of the unknown nucleus immediately after such a collision is mathematically evaluated as
Now 
because before collision the the nucleus was at rest
So
=> 
Recall that
So
=> 
substituting values
