Answer:

Explanation:
According to the law of conservation of energy, the energy of the absorbed photon must be equal to the binding energy of the electron plus the energy of the released electron:

1 eV is equal to
, so:

Solving for
and replacing the given values:

18.5164213 if you divide them both you get that number so the volceity is the number shown above.
Answer:
7kgm/s
Explanation:
Using the law of conservation of momentum which states that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision.
Let P1A and P1B be the initial momentum of the bodies A and B respectively
Let P2A and P2B be the final momentum of the bodies A and B respectively after collision.
Based on the law:
P1A+P2A = P1B + P2B
Given P1A = 5kgm/s
P2A = 0kgm/s(ball B at rest before collision)
P2A = -2.0kgm/s (negative because it moves in the negative x direction)
P2B = ?
Substituting the values in the equation gives;
5+0 = -2+P2B
5+2 = P2B
P2B = 7kgm/s
An electron is emitted in both positive and negative beta decay, although the positive one is called positron emission.
Answer:
the magnitude of first force = 3 × 5= 15 N
ANd, the magnitude of second force = 5 × 5 = 25 N
Explanation:
The computation of the magnitude of the each force is shown below:
Provided that
Ratio of forces = 3: 5
Let us assume the common factor is x
Now
first force = 3x
And, the second force = 5x
Resultant force = 35 N
The Angle between the forces = 60 degrees
Based on the above information
Resultant force i.e. F = √ F_1^2 +F_2^2 + 2 F_1F_2cos
35 = √[(3x)²+ (5x)²+ 2 (3x)(5x) cos 60°]
35 =√ 9x² + 25x² + 15x² (cos 60° = 0.5)
35 = √49 x²
x = 5
So, the magnitude of first force = 3 × 5= 15 N
ANd, the magnitude of second force = 5 × 5 = 25 N