B = 0.018 T Ans,
Since, it is moving in a circular path, thus, centripetal force will act on it i.e.
F =
where, m is the mass of the object, v is the velocity and r is the radius of circular path.
And, since a positive charge is moving, it will create magnetic force which is equal to F = qvB
where q is the charge, v is the velocity of the particle and B is the magnetic field.
Now, the two forces will be equal,
i.e.
= qvB
⇒
= qB
⇒B =
<span>putting the values, we get,
</span>
use q = 1.6 * 10^ -19
⇒ B = 0.018 T
The independent variable in this problem would be the different types of shower cleaner. The dependent variable would be the shower tiles.
The Professor's centripetal acceleration is 0.044 m/s²
Centripetal acceleration is the acceleration of an object moving in circular motion. It is usually directed towards the center of the rotation.
It is given by:
a = v²/r
where v is the velocity and r is the radius.
Given that the radius (r) = 4 m, velocity (v) = 0.419 m/s, hence:
a = v²/r = 0.419²/4 = 0.044 m/s²
The Professor's centripetal acceleration is 0.044 m/s²
Find out more at: brainly.com/question/6082363
Answer:
a)The approximate radius of the nucleus of this atom is 4.656 fermi.
b) The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527
Explanation:
= Constant for all nuclei
r = Radius of the nucleus
A = Number of nucleons
a) Given atomic number of an element = 25
Atomic mass or nucleon number = 52
The approximate radius of the nucleus of this atom is 4.656 fermi.
b) 
k=
= Coulombs constant
= charges kept at distance 'a' from each other
F = electrostatic force between charges
Force of repulsion between two protons on opposite sides of the diameter
The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527
The frequency of the
scattered photon decreases or it will be lower compare to the frequency of
incident photon. An x-ray photon scatters in one direction after a collision
and some energy is transferred to the electron as it recoils in another
direction resulting to have less energy in the scattered photon. In addition, the
frequencies will also depend on the differences of the angle at which the
scattered photon leaves the collision and this incident is called Compton Effect.
