Complete question:
At a particular instant, an electron is located at point (P) in a region of space with a uniform magnetic field that is directed vertically and has a magnitude of 3.47 mT. The electron's velocity at that instant is purely horizontal with a magnitude of 2×10⁵ m/s then how long will it take for the particle to pass through point (P) again? Give your answer in nanoseconds.
[<em>Assume that this experiment takes place in deep space so that the effect of gravity is negligible.</em>]
Answer:
The time it will take the particle to pass through point (P) again is 1.639 ns.
Explanation:
F = qvB
Also;

solving this two equations together;

where;
m is the mass of electron = 9.11 x 10⁻³¹ kg
q is the charge of electron = 1.602 x 10⁻¹⁹ C
B is the strength of the magnetic field = 3.47 x 10⁻³ T
substitute these values and solve for t

Therefore, the time it will take the particle to pass through point (P) again is 1.639 ns.
An incompressible flow field F in a 3D cartesian grid with components u,v,w:
F = u + v + w
where u,v,w are functions of x,y,z
Must satisfy:
∇·F = du/dx + dv/dy + dw/dz = 0
We have a field F defined:
F = u+v+w, u = ax+byz, v = cy+dxz
du/dx = a, dv/dy = c
Recall ∇·F = 0:
∇·F = du/dx + dv/dy + dw/dz = 0
a + c + dw/dz = 0
dw/dz = -a-c
Solve for w by separation of variables:
w = ∫(-a-c)dz
w = -az - cz + f(x,y)
f(x,y) is some undetermined function of x and y
The question states that w is not a function of x and y, therefore f(x,y) = 0...
w = -az - cz
If an atom contains 13 protons, then it has <u>13 electrons.</u>
Answer:
a) 86 atm
b) 86 atm
c) 645 m/s
Explanation:
See attachment for calculations on how i arrived at the answer