Proton: Its electrical charge is +1, its mass is 1, it's found in the nucleus
Neutron: Its electrical charge is 0, its mass is 1, it's found in the nucleus
Electron: Its electrical charge is -1, its mass is negligible, it's found outside the nucleus
Explanation:
Atoms consist of three types of particles:
- Proton: the proton is found in the nucleus of the atom, and it has a positive electric charge, equal to
(expressed in units of fundamental charge, it has a charge of +1). Its mass is 
, while its mass expressed in atomic mass units is 1 a.m.u - Neutron: the neutron is also found in the nucleus of the atom, and it has no electric charge. Its mass is similar to the mass of the proton (slightly larger), and neutrons and protons are held together in the nucleus by the presence of the strong nuclear force
- Electron: the electron orbits around the nucleus, far away from it. It has negative electric charge, opposite to that of the proton (
, or -1 in units of fundamental charge). Its mass is much lower than that of the proton, approximately 1800 times smaller, so it can be considered as negligible.
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Answer:
The answer to your question is: vf = 30 m/s
Explanation:
Data
vo = 15 m/s
a = 3.0 m/s²
t = 5 s
vf = ?
Formula
vf = vo + at
Substitution
vf = 15 + (3)(5)
vf = 15 + 15
vf = 30 m/s
Answer:
Explanation:
According to Coulomb's law, the magnitude of the force (F) between two objects with the same charge(+Q), separated a distance (d) apart, is defined as:
Here k is the Coulomb constant. The charge on each object is doubled, that is 
:
<h2>
She will find the ball at a horizontal distance of 86.4 m from landed location</h2>
Explanation:
Consider the vertical motion of ball
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 2 m/s
Acceleration, a = 9.81 m/s²
Displacement, s = 100 m
Substituting
s = ut + 0.5 at²
100 = 2 x t + 0.5 x 9.81 xt²
4.905t² + 2t - 100 = 0
t = 4.32 s or t = -4.72 s
After 4.32 seconds the ball reaches ground.
Now we need to find horizontal distance traveled by ball in 4.32 seconds.
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 20 m/s
Acceleration, a = 0 m/s²
Time, t = 4.32 s
Substituting
s = ut + 0.5 at²
s = 20 x 4.32 + 0.5 x 0 x 4.32²
s = 86.4 m
She will find the ball at a horizontal distance of 86.4 m from landed location
Hi! Your answer would be "node"
