Answer:
208 Joules
Explanation:
The radius of the circular path the charge moves, r = 26 m
The magnetic force acting on the charge particle, F = 16 N
Centripetal force,
= m·v²/r
Kinetic energy, K.E. = (1/2)·m·v²
Where;
m = The mass of the charged particle
v = The velocity of the charged particle
r = The radius of the path of the charged particle
Whereby the magnetic force acting on the charge particle = The centripetal force, we have;
F =
= m·v²/r = 16 N
(1/2) × r ×
= (1/2) × r × m·v²/r = (1/2)·m·v² = K.E.
∴ (1/2) × r ×
= (1/2) × 26 m × 16 N = = (1/2)·m·v² = K.E.
∴ 208 Joules = K.E.
The kinetic energy of an particle moving in the circular path, K.E. = 208 Joules.
Wind is primarily driven by differences in air pressure. These variations in air pressure are due to temperature differences caused by variations in solar energy received at the surface of the Earth.
Answer:
The change in potential energy is 
Explanation:
From the question we are told that
The magnitude of the uniform electric field is 
The distance traveled by the electron is 
Generally the force on this electron is mathematically represented as
Where F is the force and q is the charge on the electron which is a constant value of 
Thus


Generally the work energy theorem can be mathematically represented as

Where W is the workdone on the electron by the Electric field and
is the change in kinetic energy
Also workdone on the electron can also be represented as
Where
considering that the movement of the electron is along the x-axis
So

substituting values


Now From the law of energy conservation
Where
is the change in potential energy
Thus

Answer:
Magnitude of force the ground exerts on the plow = 263.234 N
Explanation:
Magnitude of force the ground exerts on the plow = Fground - Fplow
We are given that:
Fgound = 275 N
We will now calculate Fplow as follows:
Fplow = mass of horse * acceleration of plow
Fplow = 53 * 0.222
Fplow = 11.766 N
Now, substitute in the above equation to get magnitude of force the ground exerts on the plow as follows:
Magnitude of force the ground exerts on the plow = Fground - Fplow
Magnitude of force the ground exerts on the plow = 275 - 11.766
Magnitude of force the ground exerts on the plow = 263.234 N
Hope this helps :)