Answer:
2.1406 ×
m/sec
Explanation:
we know that energy is always conserved
so from the law of energy conservation

here V is the potential difference
we know that mass of proton = 1.67×
kg
we have given speed =50000m/sec
so potential difference 
now mass of electron =9.11×
so for electron

so the velocity of electron will be 2.1406×
m/sec
Answer:
the average force exerted by seatbelts on the passenger is 5625 N.
Explanation:
Given;
initial velocity of the car, u = 50 m/s
distance traveled by the car, s = 20 m
final velocity of the after coming to rest, v = 0
mass of the passenger, m = 90 kg
Determine the acceleration of the car as it hit the pile of dirt;
v² = u² + 2as
0 = 50² + (2 x 20)a
0 = 2500 + 40a
40a = -2500
a = -2500/40
a = -62.5 m/s²
The deceleration of the car is 62.5 m/s²
The force exerted on the passenger by the backward action of the car is calculated as follows;
F = ma
F = 90 x 62.5
F = 5625 N
Therefore, the average force exerted by seatbelts on the passenger is 5625 N.
Ionic bonds occur when electrons are donated from one atom to another. Each type of atom forms a characteristic number of covalentbonds with other atoms. An example of that is a hydrogen atom with one electron in its outer shell forms only one bond, its out most orbital becomes filled with two electrons.
the outermost protons