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.
Force = mass x acceleration
force = 2500kg x (20m/s / 10m/s)
force = 2500kg x 2m/s^2
force = 5000kg m/s^2 = 5kN
i hope this is right (^^)
<span>Answer:
So this involves right triangles. The height is always 100. Let the horizontal be x and the length of string be z.
So we have x2 + 1002 = z2. Now take its derivative in terms of time to get
2x(dx/dt) = 2z(dz/dt)
So at your specific moment z = 200, x = 100âš3 and dx/dt = +8
substituting, that makes dz/dt = 800âš3 / 200 or 4âš3.
Part 2
sin a = 100/z = 100 z-1 . Now take the derivative in terms of t to get
cos a (da./dt) = -100/ z2 (dz/dt)
So we know z = 200, which makes this a 30-60-90 triangle, therefore a=30 degrees or π/6 radians.
Substitute to get
cos (Ď€/6)(da/dt) = (-100/ 40000)(4âš3)
âš3 / 2 (da/dt) = -âš3 / 100
da/dt = -1/50 radians</span>
(a) 
The moment of inertia of a uniform-density disk is given by
where
M is the mass of the disk
R is its radius
In this problem,
M = 16 kg is the mass of the disk
R = 0.19 m is the radius
Substituting into the equation, we find
(b) 142.5 J
The rotational kinetic energy of the disk is given by
where
I is the moment of inertia
is the angular velocity
We know that the disk makes one complete rotation in T=0.2 s (so, this is the period). Therefore, its angular velocity is
And so, the rotational kinetic energy is
(c) 
The rotational angular momentum of the disk is given by
where
I is the moment of inertia
is the angular velocity
Substituting the values found in the previous parts of the problem, we find
U=6.9ms-¹
a=0.62ms-²
t=3.4s
V=?
using
v=u+at
v=6.9+(0.62×3.4)
v= 6.9+2.108
v=9.008ms-¹
